2025-10-11
Math Analysis Homework - Week 3
2025-10-11 _posts/homework/sem1
Math Analysis Homework - Week 3
Class 1
T1
Calculate left/right limits of f ( x ) = 2 1 x − 1 2 1 x + 1 ( x 0 = 0 ) \begin{gathered}
\text{Calculate left/right limits of }f(x)=\dfrac{2^{\frac{1}{x} }-1}{2^{\frac{1}{x} } +1} (x_0=0)
\end{gathered} Calculate left/right limits of f ( x ) = 2 x 1 + 1 2 x 1 − 1 ( x 0 = 0 )
x → 0 + , 1 x → + ∞ , lim x → 0 + = f ( x ) = 1 − 2 − 1 x 1 + 2 − 1 x = 1 x → 0 − , 1 x → − ∞ , 2 1 x → 0 , lim x → 0 − f ( x ) = − 1 \begin{gathered}
x\to 0^+,\dfrac{1}{x} \to +\infty,\lim_{x \to 0^+} = f(x)=\dfrac{1-2^{-\frac{1}{x}}}{1+2^{-\frac1x}}=1 \\
x\to 0^-,\dfrac{1}{x} \to -\infty,2^{\frac{1}x}\to 0,\lim_{x \to 0^-} f(x)=-1
\end{gathered} x → 0 + , x 1 → + ∞ , x → 0 + lim = f ( x ) = 1 + 2 − x 1 1 − 2 − x 1 = 1 x → 0 − , x 1 → − ∞ , 2 x 1 → 0 , x → 0 − lim f ( x ) = − 1
T2
lim x → 0 1 + x − 1 − x x \begin{gathered}
\lim_{x \to 0} \dfrac{\sqrt{ 1+x } -\sqrt{ 1-x } }{x}
\end{gathered} x → 0 lim x 1 + x − 1 − x
1 + x − 1 ∼ x 2 lim x → 0 1 + x − 1 − x x = lim x → 0 ( 1 + x − 1 ) − ( 1 − x − 1 ) x = lim x → 0 x 2 + x 2 x = 1 \begin{gathered}
\sqrt{1+x}-1\sim\dfrac{x}{2} \\
\lim_{x \to 0} \dfrac{\sqrt{ 1+x } -\sqrt{ 1-x } }{x} \\
=\lim_{x \to 0}\dfrac{(\sqrt{ 1+x }-1) -(\sqrt{ 1-x }-1) }{x} \\
=\lim_{x \to 0}\dfrac{\frac{x}{2}+\frac{x}{2}}{x} \\
=1
\end{gathered} 1 + x − 1 ∼ 2 x x → 0 lim x 1 + x − 1 − x = x → 0 lim x ( 1 + x − 1 ) − ( 1 − x − 1 ) = x → 0 lim x 2 x + 2 x = 1
T3
lim x → 1 ∑ i = 1 m x i − m x − 1 \begin{gathered}
\lim_{x \to 1} \dfrac{\sum _{i = 1} ^{m} x^i -m}{x-1}
\end{gathered} x → 1 lim x − 1 ∑ i = 1 m x i − m
let f m ( x ) = ∑ i = 1 m x i − m x − 1 lim x → 1 f 1 ( x ) = 1 f n ( x ) − f n − 1 ( x ) = x m − 1 x − 1 = ∑ i = 0 m − 1 x i ∴ lim x → 1 f n ( x ) = lim x → 1 f 1 ( x ) + ∑ i = 2 n f ( i ) − f ( i − 1 ) = 1 + ∑ i = 2 n i = n ( n + 1 ) 2 \begin{gathered}
\text{let } f_m(x)=\dfrac{\sum _{i = 1} ^{m} x^i-m}{x-1} \\
\lim_{x \to 1} f_1(x)=1 \\
f_n(x)-f_{n-1}(x)=\dfrac{x^m-1}{x-1}=\sum _{i = 0} ^{m-1} x^i \\
\therefore \lim_{x \to 1} f_n(x)=\lim_{x \to 1} f_1(x)+\sum _{i = 2} ^{n} f(i)-f(i-1) \\
=1+\sum _{i = 2} ^{n} i \\
=\dfrac{n(n+1)}{2}
\end{gathered} let f m ( x ) = x − 1 ∑ i = 1 m x i − m x → 1 lim f 1 ( x ) = 1 f n ( x ) − f n − 1 ( x ) = x − 1 x m − 1 = i = 0 ∑ m − 1 x i ∴ x → 1 lim f n ( x ) = x → 1 lim f 1 ( x ) + i = 2 ∑ n f ( i ) − f ( i − 1 ) = 1 + i = 2 ∑ n i = 2 n ( n + 1 )
T4
lim n → ∞ ∏ i = 1 n cos x 2 i \begin{gathered}
\lim_{n \to \infty} \prod_{i=1}^n \cos\dfrac{x}{2^i}
\end{gathered} n → ∞ lim i = 1 ∏ n cos 2 i x
∏ i = 1 n cos x 2 i = sin x 2 n ∏ i = 1 n cos x 2 i sin x 2 n = s i n ( x ) 2 n sin x 2 n sin x 2 n ∼ x 2 n ⟹ lim n → ∞ s i n ( x ) 2 n sin x 2 n = sin ( x ) x \begin{gathered}
\prod _{i = 1} ^{n} \cos\dfrac{x}{2^i} \\
=\dfrac{\sin\dfrac{x}{2^n}\prod _{i = 1} ^{n} \cos\dfrac{x}{2^i}}{\sin\dfrac{x}{2^n} } \\
=\dfrac{sin(x)}{2^n\sin\dfrac{x}{2^n} } \\
\sin\dfrac{x}{2^n} \sim\dfrac{x}{2^n} \\
\implies \lim_{n \to \infty} \dfrac{sin(x)}{2^n\sin\dfrac{x}{2^n} } \\
=\dfrac{\sin(x)}{x}
\end{gathered} i = 1 ∏ n cos 2 i x = sin 2 n x sin 2 n x ∏ i = 1 n cos 2 i x = 2 n sin 2 n x s in ( x ) sin 2 n x ∼ 2 n x ⟹ n → ∞ lim 2 n sin 2 n x s in ( x ) = x sin ( x )
T5
lim x → ∞ ( sin 1 x + cos 1 x ) x \begin{gathered}
\lim_{x \to \infty} (\sin\dfrac{1}{x} +\cos\dfrac{1}{x} )^x
\end{gathered} x → ∞ lim ( sin x 1 + cos x 1 ) x
lim x → ∞ ( sin 1 x + cos 1 x ) x = lim x → ∞ e x ln ( sin 1 x + cos 1 x ) = e lim x → ∞ x ln ( sin 1 x + cos 1 x ) = e lim x → ∞ x ( sin 1 x + ( cos 1 x − 1 ) ) = e lim x → ∞ x ( 1 x + ( 1 − 1 2 x 2 − 1 ) ) = e \begin{gathered}
\lim_{x \to \infty} (\sin\dfrac{1}{x} +\cos\dfrac{1}{x} )^x \\
=\lim_{x \to \infty} e^{x\ln(\sin\frac{1}{x}+\cos\frac1x)} \\
=e^{\lim_{x \to \infty} x\ln(\sin\frac{1}{x}+\cos\frac1x)} \\
=e^{\lim_{x \to \infty} x(\sin\frac{1}{x}+(\cos\frac1x-1))} \\
=e^{\lim_{x \to \infty} x(\frac{1}{x}+(1-\frac{1}{2x^2} -1))} \\
=e
\end{gathered} x → ∞ lim ( sin x 1 + cos x 1 ) x = x → ∞ lim e x l n ( s i n x 1 + c o s x 1 ) = e l i m x → ∞ x l n ( s i n x 1 + c o s x 1 ) = e l i m x → ∞ x ( s i n x 1 + ( c o s x 1 − 1 )) = e l i m x → ∞ x ( x 1 + ( 1 − 2 x 2 1 − 1 )) = e
T6
lim x → 0 x [ 1 x ] \begin{gathered}
\lim_{x \to 0} x \lbrack \dfrac{1}{x} \rbrack
\end{gathered} x → 0 lim x [ x 1 ]
{ x [ 1 x ] ∈ ( 1 − x , 1 ] lim x → 0 1 − x = lim x → 1 1 = 1 ⟹ Squeeze Theorem lim x → 0 x [ 1 x ] = 1 \begin{gathered}
\begin{cases}
x\lbrack \dfrac{1}{x} \rbrack \in (1-x,1] \\
\lim_{x \to 0} 1-x=\lim_{x \to 1} 1=1
\end{cases} \\
\stackrel{\text{Squeeze Theorem}}{\Longrightarrow}
\lim_{x \to 0} x \lbrack \dfrac{1}{x} \rbrack =1
\end{gathered} ⎩ ⎨ ⎧ x [ x 1 ] ∈ ( 1 − x , 1 ] lim x → 0 1 − x = lim x → 1 1 = 1 ⟹ Squeeze Theorem x → 0 lim x [ x 1 ] = 1
T7
lim x → ∞ ( 1 + x 3 + x ) x \begin{gathered}
\lim_{x \to \infty} (\dfrac{1+x}{3+x} )^x
\end{gathered} x → ∞ lim ( 3 + x 1 + x ) x
( 1 + x 3 + x ) x = ( 1 − 2 x + 3 ) x let t = − x + 3 2 , x = − 2 t − 3 ( 1 + 2 x + 3 ) x = ( 1 + 1 t ) − 2 t − 3 ∴ lim x → ∞ ( 1 + 2 x + 3 ) x = lim t → ∞ ( 1 + 1 t ) − 2 t = e − 2 \begin{gathered}
(\dfrac{1+x}{3+x} )^x
=(1-\dfrac{2}{x+3} )^x \\
\text{let } t=-\dfrac{x+3}{2},x=-2t-3 \\
(1+\dfrac{2}{x+3} )^x=(1+\dfrac{1}{t} )^{-2t-3} \\
\therefore \lim_{x \to \infty} (1+\dfrac{2}{x+3} )^x \\
=\lim_{t \to \infty} (1+\dfrac{1}{t} )^{-2t} \\
=e^{-2}
\end{gathered} ( 3 + x 1 + x ) x = ( 1 − x + 3 2 ) x let t = − 2 x + 3 , x = − 2 t − 3 ( 1 + x + 3 2 ) x = ( 1 + t 1 ) − 2 t − 3 ∴ x → ∞ lim ( 1 + x + 3 2 ) x = t → ∞ lim ( 1 + t 1 ) − 2 t = e − 2
T8
lim x → 0 ( a x + b x + c x 3 ) 1 x \begin{gathered}
\lim_{x \to 0} (\dfrac{a^x+b^x+c^x}{3} )^\frac{1}{x}
\end{gathered} x → 0 lim ( 3 a x + b x + c x ) x 1
lim x → 0 ( a x + b x + c x 3 ) 1 x = exp lim x → 0 ln ( a x + b x + c x 3 ) x = exp lim x → 0 ( a x + b x + c x 3 ) − 1 x a x = e x ln ( a ) = 1 + x ln ( a ) + o ( x ) ∴ = exp lim x → 0 x ln a + x ln b + x ln c x = exp ln ( a b c ) 3 = a b c 3 \begin{gathered}
\lim_{x \to 0} (\dfrac{a^x+b^x+c^x}{3} )^\frac{1}{x} \\
=\exp \lim_{x \to 0} \dfrac{\ln(\dfrac{a^x+b^x+c^x}{3} )}{x} \\
=\exp \lim_{x\to 0}\dfrac{(\dfrac{a^x+b^x+c^x}{3} )-1}{x} \\
a^x=e^{x\ln(a)}=1+x\ln(a)+o(x) \\
\therefore
=\exp \lim_{x \to 0} \dfrac{x\ln a+x\ln b+x\ln c}{x} \\
=\exp \dfrac{\ln(abc)}{3} \\
=\sqrt[ 3 ]{ abc }
\end{gathered} x → 0 lim ( 3 a x + b x + c x ) x 1 = exp x → 0 lim x ln ( 3 a x + b x + c x ) = exp x → 0 lim x ( 3 a x + b x + c x ) − 1 a x = e x l n ( a ) = 1 + x ln ( a ) + o ( x ) ∴ = exp x → 0 lim x x ln a + x ln b + x ln c = exp 3 ln ( ab c ) = 3 ab c
T9
lim x → 0 x tan 4 x sin 3 x ( 1 − cos x ) \begin{gathered}
\lim_{x \to 0} \dfrac{x\tan^4x}{\sin^3x(1-\cos x)}
\end{gathered} x → 0 lim sin 3 x ( 1 − cos x ) x tan 4 x
lim x → 0 x tan 4 x sin 3 x ( 1 − cos x ) = lim x → 0 x sin x cos 3 x ( 1 − cos x ) = lim x → 0 x 2 cos 3 x x 2 2 = 2 \begin{gathered}
\lim_{x \to 0} \dfrac{x\tan^4x}{\sin^3x(1-\cos x)} \\
=\lim_{x \to 0} \dfrac{x\sin x}{\cos^3x(1-\cos x)} \\
=\lim_{x \to 0} \dfrac{x^2}{\cos^3x\dfrac{x^2}{2} } \\
=2
\end{gathered} x → 0 lim sin 3 x ( 1 − cos x ) x tan 4 x = x → 0 lim cos 3 x ( 1 − cos x ) x sin x = x → 0 lim cos 3 x 2 x 2 x 2 = 2
T10
lim x → 0 1 + x 4 − 1 1 − cos 2 x \begin{gathered}
\lim_{x \to 0} \dfrac{\sqrt{ 1+x^4 } -1}{1-\cos^2x}
\end{gathered} x → 0 lim 1 − cos 2 x 1 + x 4 − 1
lim x → 0 1 + x 4 − 1 1 − cos 2 x = lim x → 0 x 4 2 x 2 = lim x → 0 x 2 2 = 0 \begin{gathered}
\lim_{x \to 0} \dfrac{\sqrt{ 1+x^4 } -1}{1-\cos^2x} \\
=\lim_{x \to 0} \dfrac{x^4}{2x^2} \\
=\lim_{x \to 0} \dfrac{x^2}{2} \\
=0
\end{gathered} x → 0 lim 1 − cos 2 x 1 + x 4 − 1 = x → 0 lim 2 x 2 x 4 = x → 0 lim 2 x 2 = 0
T11
solve a,b:
lim x → + ∞ ( x 2 − x + 1 − a x − b ) = 0 \begin{gathered}
\lim_{x \to +\infty} (\sqrt{ x^2-x+1 } -ax-b)=0
\end{gathered} x → + ∞ lim ( x 2 − x + 1 − a x − b ) = 0
x 2 − x + 1 = ( x − 1 2 ) 2 + 3 4 let t = x − 1 2 lim x → ∞ x 2 − x + 1 − ( x − 1 2 ) = lim x → ∞ t 2 + 3 4 − t 2 = lim x → ∞ 3 4 ( t 2 + 3 4 + t 2 ) = 0 \begin{gathered}
\sqrt{ x^2-x+1 }=\sqrt{(x-\dfrac{1}{2} )^2+\dfrac{3}{4} } \\
\text{let } t=x-\dfrac{1}{2} \\
\lim_{x \to \infty} \sqrt{ x^2-x+1 } -(x-\dfrac{1}{2} ) \\
=\lim_{x \to \infty} \sqrt{ t^2+\dfrac{3}{4} }-\sqrt{ t^2 } \\
=\lim_{x \to \infty} \dfrac{3}{4(\sqrt{t^2+\frac{3}{4}}+\sqrt{t^2})} \\
=0
\end{gathered} x 2 − x + 1 = ( x − 2 1 ) 2 + 4 3 let t = x − 2 1 x → ∞ lim x 2 − x + 1 − ( x − 2 1 ) = x → ∞ lim t 2 + 4 3 − t 2 = x → ∞ lim 4 ( t 2 + 4 3 + t 2 ) 3 = 0
T12
f ( x 0 − ) < f ( x 0 + ) ⟹ ∃ δ > 0 , ∀ x ∈ ( x 0 − δ , x 0 ) , ∀ y ∈ ( x 0 , x 0 + δ ) , f ( x ) < f ( y ) \begin{gathered}
f(x_0^-)<f(x_0^+) \implies \exists \delta>0,\forall x\in (x_0-\delta,x_0),\forall y\in (x_0,x_0+\delta),f(x)<f(y)
\end{gathered} f ( x 0 − ) < f ( x 0 + ) ⟹ ∃ δ > 0 , ∀ x ∈ ( x 0 − δ , x 0 ) , ∀ y ∈ ( x 0 , x 0 + δ ) , f ( x ) < f ( y )
lim x → x 0 − f ( x ) < lim x → x 0 + f ( x ) let ϵ = f ( x 0 + ) − f ( x 0 − ) 3 , ∃ δ = min ( δ 1 , δ 2 ) s . t . ∀ x ∈ ( x 0 − δ , x 0 ) , ∣ f ( x ) − f ( x 0 − ) ∣ < ϵ , ∀ y ∈ ( x 0 , x 0 + δ ) , ∣ f ( y ) − f ( x 0 + ) ∣ < ϵ ∴ f ( x ) < f ( x 0 − ) + ϵ < f ( x 0 + ) − ϵ < f ( y ) \begin{gathered}
\lim_{x \to x_0^-} f(x)<\lim_{x \to x_0^+} f(x) \\
\text{let } \epsilon=\dfrac{f(x_0^+)-f(x_0^-)}{3} ,\exists \delta=\min(\delta_1,\delta_2) \\ s.t.\\
\forall x\in (x_0-\delta,x_0), \vert f(x)-f(x_0^-) \vert <\epsilon, \\
\forall y\in (x_0,x_0+\delta), \vert f(y)-f(x_0^+) \vert <\epsilon \\
\therefore f(x)<f(x_0^-)+\epsilon<f(x_0^+)-\epsilon<f(y)
\end{gathered} x → x 0 − lim f ( x ) < x → x 0 + lim f ( x ) let ϵ = 3 f ( x 0 + ) − f ( x 0 − ) , ∃ δ = min ( δ 1 , δ 2 ) s . t . ∀ x ∈ ( x 0 − δ , x 0 ) , ∣ f ( x ) − f ( x 0 − ) ∣ < ϵ , ∀ y ∈ ( x 0 , x 0 + δ ) , ∣ f ( y ) − f ( x 0 + ) ∣ < ϵ ∴ f ( x ) < f ( x 0 − ) + ϵ < f ( x 0 + ) − ϵ < f ( y )
T13
f is periodic function , lim x → ∞ f ( x ) = 0 ⟹ f ( x ) = 0 \begin{gathered}
f \text{ is periodic function} ,\lim_{x \to \infty} f(x)=0 \\
\implies f(x)=0
\end{gathered} f is periodic function , x → ∞ lim f ( x ) = 0 ⟹ f ( x ) = 0
let T is a period of f ∀ x 0 , ϵ , ∃ X > x 0 s . t . x > X ⟹ f ( x ) < ϵ let n = [ X − x 0 T + 100 ] ⟹ f ( x 0 ) = f ( x 0 + n T ) < ϵ ⟹ lim x → x 0 f ( x ) = 0 ⟹ f ( x ) = 0 \begin{gathered}
\text{let } T \text{ is a period of } f \\
\forall x_0,\epsilon,
\exists X>x_0 \ s.t.\
x>X \implies f(x)<\epsilon \\
\text{let } n=\lbrack \dfrac{X-x_0}{T} +100 \rbrack \\
\implies f(x_0)=f(x_0+nT)<\epsilon \\
\implies \lim_{x \to x_0} f(x)=0 \\
\implies f(x)=0
\end{gathered} let T is a period of f ∀ x 0 , ϵ , ∃ X > x 0 s . t . x > X ⟹ f ( x ) < ϵ let n = [ T X − x 0 + 100 ] ⟹ f ( x 0 ) = f ( x 0 + n T ) < ϵ ⟹ x → x 0 lim f ( x ) = 0 ⟹ f ( x ) = 0
T14
{ f ( x ) , x ∈ ( 0 , 1 ) x → 0 + ⟹ f ( x ) = o ( 1 ) f ( x ) − f ( x 2 ) = o ( x ) ⟹ x → 0 + , f ( x ) = o ( x ) \begin{gathered}
\begin{cases}
f(x),x\in(0,1) \\
x\to 0^+ \implies f(x)=o(1) \\
f(x)-f(\dfrac{x}{2} )=o(x)
\end{cases}
\implies x\to 0^+ ,f(x)=o(x)
\end{gathered} ⎩ ⎨ ⎧ f ( x ) , x ∈ ( 0 , 1 ) x → 0 + ⟹ f ( x ) = o ( 1 ) f ( x ) − f ( 2 x ) = o ( x ) ⟹ x → 0 + , f ( x ) = o ( x )
∀ ϵ , let ϵ 1 = ϵ 8 ∃ δ s . t . x < δ ⟹ ∣ f ( x ) − f ( x 2 ) ∣ < ϵ 1 x ∴ ∣ f ( x ) ∣ = ∣ f ( x 2 n ) + ∑ i = 0 n − 1 ( f ( x 2 i ) − f ( x 2 i + 1 ) ) ∣ ≤ ∣ f ( x 2 n ) ∣ + ∑ i = 0 n − 1 ∣ f ( x 2 i ) − f ( x 2 i + 1 ) ∣ < ∣ f ( x 2 n ) ∣ + ∑ i = 0 n − 1 ϵ 1 x 2 i < ∣ f ( x 2 n ) ∣ + 2 ϵ 1 x ∴ ∣ f ( x ) ∣ = lim n → ∞ ∣ f ( x ) ∣ ≤ lim n → ∞ ( ∣ f ( x 2 n ) ∣ + 2 ϵ 1 x ) = 0 + 2 ϵ 1 x = 2 ϵ 1 x < ϵ x ⟹ x → 0 + ⟹ f ( x ) = o ( x ) \begin{gathered}
\forall \epsilon, \\
\text{let }\epsilon_1=\dfrac{\epsilon}{8}\\
\exists \delta \ s.t.\
x<\delta \implies
\vert f(x)-f(\dfrac{x}{2} )\vert <\epsilon_1x \\
\therefore
\vert f(x)\vert = \vert f(\dfrac{x}{2^n} )+\sum _{i = 0} ^{n-1} ( f(\dfrac{x}{2^i} )-f(\dfrac{x}{2^{i+1}} ) )\vert \\
\le \vert f(\dfrac{x}{2^n} )\vert + \sum _{i = 0} ^{n-1} \vert f(\dfrac{x}{2^i} )-f(\dfrac{x}{2^{i+1}} ) \vert \\
<\vert f(\dfrac{x}{2^n} )\vert + \sum _{i = 0} ^{n-1} \dfrac{\epsilon_1x}{2^i} \\
<\vert f(\dfrac{x}{2^n} )\vert + 2\epsilon_1x \\
\therefore \vert f(x)\vert =\lim_{n \to \infty} \vert f(x)\vert \\
\le \lim_{n \to \infty} (\vert f(\dfrac{x}{2^n} )\vert +2\epsilon_1x) \\
= 0 + 2\epsilon_1x \\
= 2\epsilon_1x \\
< \epsilon x \\
\implies x\to 0^+ \implies f(x)=o(x)
\end{gathered} ∀ ϵ , let ϵ 1 = 8 ϵ ∃ δ s . t . x < δ ⟹ ∣ f ( x ) − f ( 2 x ) ∣ < ϵ 1 x ∴ ∣ f ( x ) ∣ = ∣ f ( 2 n x ) + i = 0 ∑ n − 1 ( f ( 2 i x ) − f ( 2 i + 1 x )) ∣ ≤ ∣ f ( 2 n x ) ∣ + i = 0 ∑ n − 1 ∣ f ( 2 i x ) − f ( 2 i + 1 x ) ∣ < ∣ f ( 2 n x ) ∣ + i = 0 ∑ n − 1 2 i ϵ 1 x < ∣ f ( 2 n x ) ∣ + 2 ϵ 1 x ∴ ∣ f ( x ) ∣ = n → ∞ lim ∣ f ( x ) ∣ ≤ n → ∞ lim ( ∣ f ( 2 n x ) ∣ + 2 ϵ 1 x ) = 0 + 2 ϵ 1 x = 2 ϵ 1 x < ϵ x ⟹ x → 0 + ⟹ f ( x ) = o ( x )
T15
{ a , b > 1 , f ( x ) is bounded in N ∗ ( 0 ) f ( a x ) = b f ( x ) ⟹ lim x → 0 f ( x ) = f ( 0 ) \begin{gathered}
\begin{cases}
a,b>1,f(x) \text{ is bounded in } N^*(0) \\
f(ax)=bf(x)
\end{cases} \\
\implies \lim_{x \to 0} f(x)=f(0)
\end{gathered} { a , b > 1 , f ( x ) is bounded in N ∗ ( 0 ) f ( a x ) = b f ( x ) ⟹ x → 0 lim f ( x ) = f ( 0 )
f ( a x ) = b f ( x ) ⟺ f ( x a ) = f ( x ) b a , b > 1 , f ( x ) is bounded in N ∗ ( 0 ) ⟹ ∃ M , δ 0 , ∣ x ∣ < δ 0 ⟹ ∣ f ( x ) ∣ < M ∀ ϵ , let δ = δ 0 a n , n = log b ( M ϵ ) + 1 ⟹ x < δ ⟹ f ( x ) = f ( a n x ) b n , ∣ a n x ∣ < δ 0 ⟹ f ( a n x ) < M ⟹ f ( x ) < M b n < ϵ ⟹ lim x → 0 f ( x ) = 0 f ( a 0 ) = b f ( 0 ) ⟹ f ( 0 ) = 0 ∴ lim x → 0 f ( x ) = 0 \begin{gathered}
f(ax)=bf(x) \\
\iff f(\dfrac{x}{a} )=\dfrac{f(x)}{b} \\
a,b>1,f(x) \text{ is bounded in } N^*(0) \\
\implies \exists M,\delta_0,\vert x \vert <\delta_0 \implies \vert f(x) \vert <M \\
\forall \epsilon,\text{let }\delta=\dfrac{\delta_0}{a^n},n=\log_b(\frac{M}{\epsilon})+1 \\
\implies x<\delta \implies f(x)=\dfrac{f(a^nx)}{b^n},\vert a^nx \vert < \delta_0 \\
\implies f(a^nx)< M \\
\implies f(x)<\dfrac{M}{b^n} <\epsilon \\
\implies \lim_{x \to _0} f(x)=0 \\
f(a0)=bf(0) \implies f(0)=0 \\
\therefore \lim_{x \to 0} f(x)=0
\end{gathered} f ( a x ) = b f ( x ) ⟺ f ( a x ) = b f ( x ) a , b > 1 , f ( x ) is bounded in N ∗ ( 0 ) ⟹ ∃ M , δ 0 , ∣ x ∣ < δ 0 ⟹ ∣ f ( x ) ∣ < M ∀ ϵ , let δ = a n δ 0 , n = log b ( ϵ M ) + 1 ⟹ x < δ ⟹ f ( x ) = b n f ( a n x ) , ∣ a n x ∣ < δ 0 ⟹ f ( a n x ) < M ⟹ f ( x ) < b n M < ϵ ⟹ x → 0 lim f ( x ) = 0 f ( a 0 ) = b f ( 0 ) ⟹ f ( 0 ) = 0 ∴ x → 0 lim f ( x ) = 0
Class 2
T1
solve a,b such that
f ( x ) = lim n → ∞ x 2 n − 1 + a x 2 + b x x 2 n + 1 is continuous \begin{gathered}
f(x)=\lim_{n \to \infty} \dfrac{x^{2n-1}+ax^2+bx}{x^{2n}+1} \text{ is continuous}
\end{gathered} f ( x ) = n → ∞ lim x 2 n + 1 x 2 n − 1 + a x 2 + b x is continuous
x > 1 ⟹ f ( x ) = 1 x ⟹ lim x → 1 + f ( x ) = 1 x ∈ ( − 1 , 1 ) ⟹ f ( x ) = a x 2 + b x ⟹ lim x → 1 − f ( x ) = a x 2 + b x ∴ f ( 1 ) = 1 + a + b 2 = 1 = a + b x < − 1 ⟹ f ( x ) = 1 x ⟹ lim x → − 1 − f ( x ) = − 1 lim x → − 1 − f ( x ) = lim x → − 1 + f ( x ) = f ( − 1 ) ⟹ − 1 = a − b = − 1 + a − b 2 ∴ { a = 0 b = 1 \begin{gathered}
x>1 \implies f(x)=\dfrac{1}{x} \\
\implies \lim_{x \to 1^+} f(x)=1 \\
x\in (-1,1) \implies f(x)=ax^2+bx \\
\implies \lim_{x \to 1^-} f(x)=ax^2+bx \\
\therefore f(1)=\dfrac{1+a+b}{2} =1=a+b \\
x<-1 \implies f(x)=\dfrac{1}{x} \\
\implies \lim_{x \to -1^-} f(x)=-1 \\
\lim_{x \to -1^-} f(x)=\lim_{x \to -1^+} f(x)=f(-1) \\
\implies -1=a-b=\dfrac{-1+a-b}{2} \\
\therefore
\begin{cases}
a=0 \\
b=1
\end{cases}
\end{gathered} x > 1 ⟹ f ( x ) = x 1 ⟹ x → 1 + lim f ( x ) = 1 x ∈ ( − 1 , 1 ) ⟹ f ( x ) = a x 2 + b x ⟹ x → 1 − lim f ( x ) = a x 2 + b x ∴ f ( 1 ) = 2 1 + a + b = 1 = a + b x < − 1 ⟹ f ( x ) = x 1 ⟹ x → − 1 − lim f ( x ) = − 1 x → − 1 − lim f ( x ) = x → − 1 + lim f ( x ) = f ( − 1 ) ⟹ − 1 = a − b = 2 − 1 + a − b ∴ { a = 0 b = 1
T2
f ( x ) = { x a sin 1 x , x > 0 e x + b , x ≤ 0 \begin{gathered}
f(x)=\begin{cases}
x^a\sin\dfrac{1}{x} ,x>0 \\
e^x+b,x\le 0
\end{cases}
\end{gathered} f ( x ) = ⎩ ⎨ ⎧ x a sin x 1 , x > 0 e x + b , x ≤ 0 survey continuouity of f ( 0 ) f(0) f ( 0 )
f ( 0 ) = b + 1 lim x → 0 − f ( x ) = b + 1 if a > 0 , lim x → 0 + f ( x ) = x a sin 1 x ∈ ( − x a , x a ) lim x → 0 + f ( x ) = 0 if a ≤ 0 , lim x → 0 + f ( x ) = lim x → 0 + x a sin 1 x not exists ∴ a > 0 , b = − 1 : f ( x ) is continuous at x = 0 else f ( x ) is discontinuous at x = 0 \begin{gathered}
f(0)=b+1 \\
\lim_{x \to 0^-} f(x)=b+1 \\
\text{if } a> 0,\lim_{x \to 0^+} f(x)=x^a\sin\dfrac{1}{x} \in(-x^a,x^a) \\
\lim_{x \to 0^+} f(x)=0 \\
\text{if } a\le 0,\lim_{x \to 0^+} f(x)=\lim_{x \to 0^+} x^a\sin\dfrac{1}{x} \text{ not exists}
\\
\therefore a>0,b=-1: f(x) \text{ is continuous at } x=0 \\
\text{else } f(x) \text{ is discontinuous at } x=0
\end{gathered} f ( 0 ) = b + 1 x → 0 − lim f ( x ) = b + 1 if a > 0 , x → 0 + lim f ( x ) = x a sin x 1 ∈ ( − x a , x a ) x → 0 + lim f ( x ) = 0 if a ≤ 0 , x → 0 + lim f ( x ) = x → 0 + lim x a sin x 1 not exists ∴ a > 0 , b = − 1 : f ( x ) is continuous at x = 0 else f ( x ) is discontinuous at x = 0
T3
lim x → 0 ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1 x \begin{gathered}
\lim_{x \to 0} \dfrac{(1+x)(1+2x)(1+3x)-1}{x}
\end{gathered} x → 0 lim x ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1
( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1 = 6 x + o ( x ) ⟹ lim x → 0 ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1 x = lim x → 0 6 x + o ( x ) x = 6 \begin{gathered}
(1+x)(1+2x)(1+3x)-1=6x+o(x) \\
\implies \lim_{x \to 0} \dfrac{(1+x)(1+2x)(1+3x)-1}{x} \\
=\lim_{x \to 0} \dfrac{6x+o(x)}{x} \\
=6
\end{gathered} ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1 = 6 x + o ( x ) ⟹ x → 0 lim x ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) − 1 = x → 0 lim x 6 x + o ( x ) = 6
T4
lim x → 1 x m − 1 x n − 1 , m , n ∈ N ∗ \begin{gathered}
\lim_{x \to 1} \dfrac{\sqrt[ m ]{ x } -1}{\sqrt[ n ]{ x } -1} ,m,n\in N^*
\end{gathered} x → 1 lim n x − 1 m x − 1 , m , n ∈ N ∗
1 + ( x − 1 ) n − 1 = x − 1 n + o ( x − 1 ) ⟹ lim x → 1 x m − 1 x n − 1 = x − 1 m + o ( x − 1 ) x − 1 n + o ( x − 1 ) = n m \begin{gathered}
\sqrt[n]{ 1+(x-1) } -1 \\
=\dfrac{x-1}{n} +o(x-1) \\
\implies \lim_{x \to 1} \dfrac{\sqrt[ m ]{ x } -1}{\sqrt[ n ]{ x } -1} =\dfrac{\frac{x-1}{m}+o(x-1)}{\frac{x-1}{n}+o(x-1)} =\dfrac{n}{m}
\end{gathered} n 1 + ( x − 1 ) − 1 = n x − 1 + o ( x − 1 ) ⟹ x → 1 lim n x − 1 m x − 1 = n x − 1 + o ( x − 1 ) m x − 1 + o ( x − 1 ) = m n
T5
lim x → 1 ( 1 − x ) ( 1 − x 3 ) … ( 1 − x n ) ( 1 − x ) n − 1 , n ∈ N + \begin{gathered}
\lim_{x \to 1} \dfrac{(1-\sqrt x)(1-\sqrt[ 3 ]{ x } )\ldots (1-\sqrt[n]{ x } )}{(1-x)^{n-1}},n\in N_{+}
\end{gathered} x → 1 lim ( 1 − x ) n − 1 ( 1 − x ) ( 1 − 3 x ) … ( 1 − n x ) , n ∈ N +
1 − x k = 1 − 1 + ( x − 1 ) k = − x − 1 k + o ( x − 1 ) \begin{gathered}
1-\sqrt[k]{ x }=1-\sqrt[k]{ 1+(x-1) }=-\dfrac{x-1}{k}+o(x-1)
\end{gathered} 1 − k x = 1 − k 1 + ( x − 1 ) = − k x − 1 + o ( x − 1 ) lim x → 1 ∏ i = 2 n ( 1 − x i ) ( 1 − x ) n − 1 = lim x → 1 ∏ i = 2 n 1 − x i ( 1 − x ) n − 1 = 1 n ! \begin{gathered}
\lim_{x \to 1} \dfrac{\prod _{i = 2} ^{n} (1-\sqrt[ i ]{ x } ) }{(1-x)^{n-1}} \\
=\lim_{x \to 1} \dfrac{\prod _{i = 2} ^{n} \dfrac{1-x}{i} }{(1-x)^{n-1}} \\
=\dfrac{1}{n!}
\end{gathered} x → 1 lim ( 1 − x ) n − 1 ∏ i = 2 n ( 1 − i x ) = x → 1 lim ( 1 − x ) n − 1 ∏ i = 2 n i 1 − x = n ! 1
T6
lim x → π 4 ( tan x ) tan 2 x \begin{gathered}
\lim_{x \to \frac{\pi}{4} } (\tan x)^{\tan 2x}
\end{gathered} x → 4 π lim ( tan x ) t a n 2 x
lim x → π 4 ( tan x ) tan 2 x = lim x → π 4 exp ( tan 2 x ln ( tan x ) ) = lim x → π 4 exp ( 2 tan x 1 − tan 2 x ( tan x − 1 ) ) = lim x → π 4 exp − 2 tan x 1 + tan x = 1 e \begin{gathered}
\lim_{x \to \frac{\pi}{4} } (\tan x)^{\tan 2x}= \\
\lim_{x \to \frac{\pi}{4} } \exp (\tan 2x\ln (\tan x)) \\
=\lim_{x \to \frac{\pi}{4} } \exp (2\dfrac{\tan x}{1-\tan^2 x} (\tan x-1)) \\
=\lim_{x \to \frac{\pi}{4} } \exp -\dfrac{2\tan x}{1+\tan x} \\
=\dfrac{1}{e}
\end{gathered} x → 4 π lim ( tan x ) t a n 2 x = x → 4 π lim exp ( tan 2 x ln ( tan x )) = x → 4 π lim exp ( 2 1 − tan 2 x tan x ( tan x − 1 )) = x → 4 π lim exp − 1 + tan x 2 tan x = e 1
T7
lim x → 0 ( 2 e x 1 + x − 1 ) 1 + x 2 x \begin{gathered}
\lim_{x \to 0} (2e^{\frac{x}{1+x}}-1)^{\frac{1+x^2}x}
\end{gathered} x → 0 lim ( 2 e 1 + x x − 1 ) x 1 + x 2
lim x → 0 ( 2 e x 1 + x − 1 ) 1 + x 2 x = lim x → 0 exp 1 + x 2 x ln ( 2 e x 1 + x − 1 ) = lim x → 0 exp 2 1 + x 2 x ( e x 1 + x − 1 ) = lim x → 0 exp 2 1 + x 2 x x 1 + x = e 2 \begin{gathered}
\lim_{x \to 0} (2e^{\frac{x}{1+x}}-1)^{\frac{1+x^2}x} \\
=\lim_{x \to 0} \exp \dfrac{1+x^2}{x} \ln(2e^{\frac{x}{1+x}}-1) \\
=\lim_{x \to 0} \exp 2\dfrac{1+x^2}{x} (e^{\frac{x}{1+x}}-1) \\
=\lim_{x \to 0} \exp 2\dfrac{1+x^2}{x} \dfrac{x}{1+x} \\
=e^2
\end{gathered} x → 0 lim ( 2 e 1 + x x − 1 ) x 1 + x 2 = x → 0 lim exp x 1 + x 2 ln ( 2 e 1 + x x − 1 ) = x → 0 lim exp 2 x 1 + x 2 ( e 1 + x x − 1 ) = x → 0 lim exp 2 x 1 + x 2 1 + x x = e 2
T8
∣ x ∣ < 1 , lim n → ∞ ( 1 + ∑ i = 1 n x i n ) n \begin{gathered}
\vert x \vert <1,\lim_{n \to \infty} {\left( 1+\dfrac{\sum _{i = 1} ^{n} x^i}{n} \right)} ^n
\end{gathered} ∣ x ∣ < 1 , n → ∞ lim ( 1 + n ∑ i = 1 n x i ) n
∑ i = 1 n x i = x n + 1 − x x − 1 lim n → ∞ ( 1 + ∑ i = 1 n x i n ) n = lim n → ∞ ( 1 + x n + 1 − x n ( x − 1 ) ) n = lim n → ∞ exp n ln ( 1 + x n + 1 − x n ( x − 1 ) ) = exp lim n → ∞ n x n + 1 − x n ( x − 1 ) = exp − x x − 1 = e x 1 − x \begin{gathered}
\sum _{i = 1} ^{n} x^i=\dfrac{x^{n+1}-x}{x-1} \\
\lim_{n \to \infty} {\left( 1+\dfrac{\sum _{i = 1} ^{n} x^i}{n} \right)} ^n \\
= \lim_{n \to \infty} {\left( 1+\dfrac{x^{n+1}-x}{n(x-1)} \right)} ^n \\
=\lim_{n\to \infty} \exp n\ln {\left( 1+\dfrac{x^{n+1}-x}{n(x-1)} \right)} \\
=\exp \lim_{n \to \infty} n \dfrac{x^{n+1}-x}{n(x-1)} \\
=\exp \dfrac{-x}{x-1} \\
=e^{\frac{x}{1-x} }
\end{gathered} i = 1 ∑ n x i = x − 1 x n + 1 − x n → ∞ lim ( 1 + n ∑ i = 1 n x i ) n = n → ∞ lim ( 1 + n ( x − 1 ) x n + 1 − x ) n = n → ∞ lim exp n ln ( 1 + n ( x − 1 ) x n + 1 − x ) = exp n → ∞ lim n n ( x − 1 ) x n + 1 − x = exp x − 1 − x = e 1 − x x
T9
f ∈ C ( 0 , + ∞ ) , f ( x 2 ) = f ( x ) ⟹ ∃ c , f ( x ) = c \begin{gathered}
f\in C(0,+\infty),f(x^2)=f(x) \implies \exists c,f(x)=c
\end{gathered} f ∈ C ( 0 , + ∞ ) , f ( x 2 ) = f ( x ) ⟹ ∃ c , f ( x ) = c
∀ x 0 , let a n = x 0 1 2 n − 1 f ( a n ) = f ( a n − 1 ) = f ( a n − 1 ) , f ( a 1 ) = f ( x 0 ) ⟹ f ( a n ) = f ( x 0 ) lim n → ∞ a n = 1 ⟹ f ( 1 ) = lim x → 1 f ( x ) = lim n → ∞ f ( a n ) = lim n → ∞ f ( x 0 ) = f ( x 0 ) ⟹ ∀ x 0 , f ( x 0 ) = f ( 1 ) ⟹ f is constant function \begin{gathered}
\forall x_0,\text{let } a_n=x_0^{\frac{1}{2^{n-1}}} \\
f(a_n)=f(\sqrt{a_{n-1}})=f(a_{n-1}),f(a_1)=f(x_0) \\
\implies f(a_n)=f(x_0) \\
\lim_{n \to \infty} a_n=1 \\
\implies
f(1)=\lim_{x \to 1} f(x)=\lim_{n \to \infty} f(a_n) \\
=\lim_{n \to \infty} f(x_0)=f(x_0) \\
\implies \forall x_0,f(x_0)=f(1) \\
\implies f \text{ is constant function}
\end{gathered} ∀ x 0 , let a n = x 0 2 n − 1 1 f ( a n ) = f ( a n − 1 ) = f ( a n − 1 ) , f ( a 1 ) = f ( x 0 ) ⟹ f ( a n ) = f ( x 0 ) n → ∞ lim a n = 1 ⟹ f ( 1 ) = x → 1 lim f ( x ) = n → ∞ lim f ( a n ) = n → ∞ lim f ( x 0 ) = f ( x 0 ) ⟹ ∀ x 0 , f ( x 0 ) = f ( 1 ) ⟹ f is constant function
T10
f ( x + y ) = f ( x ) + f ( y ) , f ( x ) is continuous at x = 0 ⟹ f ∈ C ( R ) \begin{gathered}
f(x+y)=f(x)+f(y),f(x) \text{ is continuous at } x=0 \\
\implies f\in C(R)
\end{gathered} f ( x + y ) = f ( x ) + f ( y ) , f ( x ) is continuous at x = 0 ⟹ f ∈ C ( R )
f ( 0 + 0 ) = f ( 0 ) + f ( 0 ) ⟹ f ( 0 ) = 0 f ( x ) is continuous at x = 0 ⟹ ∀ ϵ 0 , ∃ δ 0 , ∣ x ∣ < δ 0 ⟹ ∣ f ( x ) − f ( 0 ) ∣ < ϵ 0 ∴ ∀ x 0 , ∀ ϵ , let ϵ 0 = ϵ 2 ⟹ δ : = δ 0 ∣ x − x 0 ∣ < δ ⟹ ∣ f ( x ) − f ( x 0 ) ∣ = ∣ f ( x 0 + ( x − x 0 ) ) − f ( x 0 ) ∣ = ∣ f ( x 0 ) + f ( x − x 0 ) − f ( x 0 ) ∣ = ∣ f ( x − x 0 ) ∣ < ϵ 0 < ϵ ∴ ∀ x 0 , lim x → x 0 f ( x ) = f ( x 0 ) ⟹ f ∈ C ( R ) \begin{gathered}
f(0+0)=f(0)+f(0)\implies f(0)=0 \\
f(x) \text{ is continuous at } x=0 \\
\implies \forall \epsilon_0,\exists \delta_0,\vert x \vert<\delta_0 \implies \vert f(x)-f(0) \vert <\epsilon_0 \\
\therefore \forall x_0, \\
\forall \epsilon,\text{let } \epsilon_0=\dfrac{\epsilon}{2} \implies \delta:=\delta_0 \\
\vert x-x_0 \vert <\delta \implies \\
\vert f(x)-f(x_0) \vert =\vert f(x_0+(x-x_0))-f(x_0) \vert \\
=\vert f(x_0)+f(x-x_0)-f(x_0) \vert \\
=\vert f(x-x_0) \vert <\epsilon_0<\epsilon \\
\therefore \forall x_0, \lim_{x \to x_0} f(x)=f(x_0) \\
\implies f\in C(R)
\end{gathered} f ( 0 + 0 ) = f ( 0 ) + f ( 0 ) ⟹ f ( 0 ) = 0 f ( x ) is continuous at x = 0 ⟹ ∀ ϵ 0 , ∃ δ 0 , ∣ x ∣ < δ 0 ⟹ ∣ f ( x ) − f ( 0 ) ∣ < ϵ 0 ∴ ∀ x 0 , ∀ ϵ , let ϵ 0 = 2 ϵ ⟹ δ := δ 0 ∣ x − x 0 ∣ < δ ⟹ ∣ f ( x ) − f ( x 0 ) ∣ = ∣ f ( x 0 + ( x − x 0 )) − f ( x 0 ) ∣ = ∣ f ( x 0 ) + f ( x − x 0 ) − f ( x 0 ) ∣ = ∣ f ( x − x 0 ) ∣ < ϵ 0 < ϵ ∴ ∀ x 0 , x → x 0 lim f ( x ) = f ( x 0 ) ⟹ f ∈ C ( R )
T11
f : [ 0 , + ∞ ) ⟹ R , f ( 2 x ) = f ( x ) cos ( x ) , f ( x ) is continuous at x = 0 ⟹ f ( x ) = f ( 0 ) sin ( x ) x , x ∈ [ 0 , + ∞ ) \begin{gathered}
f:[0,+\infty) \implies R,f(2x)=f(x)\cos(x),f(x) \text{ is continuous at } x=0 \\
\implies f(x)=f(0)\dfrac{\sin(x)}{x} ,x\in [0,+\infty)
\end{gathered} f : [ 0 , + ∞ ) ⟹ R , f ( 2 x ) = f ( x ) cos ( x ) , f ( x ) is continuous at x = 0 ⟹ f ( x ) = f ( 0 ) x sin ( x ) , x ∈ [ 0 , + ∞ )
∀ x 0 , let a n = x 0 2 n − 1 , x 0 = a 1 f ( a n ) = f ( a n + 1 ) cos ( a n + 1 ) ⟹ f ( x 0 ) = f ( a 1 ) = f ( a n ) ∏ i = 2 n cos ( a i ) = f ( a n ) ∏ i = 1 n − 1 cos x 0 2 i = f ( a n ) sin ( x 0 ) 2 n − 1 sin x 0 2 n − 1 ⟹ lim n → ∞ f ( x 0 ) = ( lim n → ∞ f ( a n ) ) sin ( x 0 ) lim n → ∞ 2 n − 1 sin x 0 2 n − 1 = f ( 0 ) sin ( x 0 ) x 0 ∴ f ( x ) = f ( 0 ) sin ( x ) x \begin{gathered}
\forall x_0, \\
\text{let } a_n=\dfrac{x_0}{2^{n-1}},x_0=a_1 \\
f(a_n)=f(a_{n+1})\cos(a_{n+1}) \\
\implies f(x_0)=f(a_1)=f(a_n)\prod _{i = 2} ^{n} \cos(a_i) \\
=f(a_n)\prod_{i=1}^{n-1}\cos \dfrac{x_0}{2^i} \\
=f(a_n)\dfrac{\sin(x_0)}{2^{n-1}\sin\frac{x_0}{2^{n-1}}} \\
\stackrel{\lim_{n \to \infty} }{\Longrightarrow}
f(x_0)=(\lim_{n \to \infty} f(a_n))\dfrac{\sin(x_0)}{\lim_{n \to \infty} 2^{n-1}\sin\frac{x_0}{2^{n-1}}} \\
=f(0)\dfrac{\sin(x_0)}{x_0} \\
\therefore f(x)=f(0)\dfrac{\sin(x)}{x}
\end{gathered} ∀ x 0 , let a n = 2 n − 1 x 0 , x 0 = a 1 f ( a n ) = f ( a n + 1 ) cos ( a n + 1 ) ⟹ f ( x 0 ) = f ( a 1 ) = f ( a n ) i = 2 ∏ n cos ( a i ) = f ( a n ) i = 1 ∏ n − 1 cos 2 i x 0 = f ( a n ) 2 n − 1 sin 2 n − 1 x 0 sin ( x 0 ) ⟹ l i m n → ∞ f ( x 0 ) = ( n → ∞ lim f ( a n )) lim n → ∞ 2 n − 1 sin 2 n − 1 x 0 sin ( x 0 ) = f ( 0 ) x 0 sin ( x 0 ) ∴ f ( x ) = f ( 0 ) x sin ( x )