Math Analysis Homework - Week 6
Class 1
T1
analysis the function's convexity and inflection point
f ( x ) = 1 − x 2 1 + x \begin{gathered}
f(x)=\dfrac{1-x^2}{1+x}
\end{gathered} f ( x ) = 1 + x 1 − x 2
f ( x ) = 1 − x ( x ≠ − 1 ) f ′ ′ ( x ) = 0 ⟹ convex in ( − ∞ , − 1 ) , ( − 1 , + ∞ ) ∀ x ≠ − 1 , x is a inflection point \begin{gathered}
f(x)=1-x (x\ne -1) \\
f''(x)=0 \\
\implies \text{convex in } (-\infty,-1),(-1,+\infty) \\
\forall x \ne -1,x \text{ is a inflection point}
\end{gathered} f ( x ) = 1 − x ( x = − 1 ) f ′′ ( x ) = 0 ⟹ convex in ( − ∞ , − 1 ) , ( − 1 , + ∞ ) ∀ x = − 1 , x is a inflection point
T2
a , b > 0 ⟹ ( a + b ) ln a + b 2 ≤ a ln a + b ln b \begin{gathered}
a,b>0 \implies \\
(a+b)\ln \dfrac{a+b}{2} \le a\ln a+b\ln b
\end{gathered} a , b > 0 ⟹ ( a + b ) ln 2 a + b ≤ a ln a + b ln b
f ( x ) = x ln x f ′ ′ ( x ) = 1 x ≥ 0 ⟹ f is convex in ( 0 , + ∞ ) ⟹ f ( a ) + f ( b ) ≥ 2 f ( a + b 2 ) ⟹ ( a + b ) ln a + b 2 ≤ a ln a + b ln b \begin{gathered}
f(x)=x\ln x \\
f''(x)=\dfrac{1}{x} \ge 0 \implies f \text{ is convex in } (0,+\infty) \\
\implies f(a)+f(b)\ge 2f(\dfrac{a+b}{2} ) \\
\implies (a+b)\ln \dfrac{a+b}{2} \le a\ln a+b\ln b
\end{gathered} f ( x ) = x ln x f ′′ ( x ) = x 1 ≥ 0 ⟹ f is convex in ( 0 , + ∞ ) ⟹ f ( a ) + f ( b ) ≥ 2 f ( 2 a + b ) ⟹ ( a + b ) ln 2 a + b ≤ a ln a + b ln b
T3
p , q > 0 ; a , b ≥ 0 ⟹ ( a p ) p ( b q ) q ≤ ( a + b p + q ) p + q \begin{gathered}
p,q>0;a,b\ge 0 \implies \\
(\dfrac{a}{p} )^p(\dfrac{b}{q} )^q\le (\dfrac{a+b}{p+q} )^{p+q}
\end{gathered} p , q > 0 ; a , b ≥ 0 ⟹ ( p a ) p ( q b ) q ≤ ( p + q a + b ) p + q
a b = 0 ab=0 ab = 0 a , b ≠ 0 a,b\ne 0 a , b = 0
( a p ) p ( b q ) q ≤ ( a + b p + q ) p + q ⟺ p ln a − p ln p + q ln b − q ln q ≤ ( p + q ) ( ln ( a + b ) − ln ( p + q ) ) ⟺ p ( ln a − ln ( a + b ) + ln ( p + q ) − ln p ) + q ( ln b − ln ( a + b ) + ln ( p + q ) − ln q ) ≤ 0 ⇔ let x = p p + q , y = a a + b x ( ln y − ln x ) + ( 1 − x ) ( ln ( 1 − y ) − ln ( 1 − x ) ) ≤ 0 f ( y ) = x ln y + ( 1 − x ) ln ( 1 − y ) f ′ ( y ) = x y − 1 − x 1 − y = x − y y ( 1 − y ) ⟹ sign f ′ ( y ) = − sign ( y − x ) , f ( x ) is a maximum ⟹ f ( y ) ≤ f ( y ) = 0 Q.E.D \begin{gathered}
(\dfrac{a}{p} )^p(\dfrac{b}{q} )^q\le (\dfrac{a+b}{p+q} )^{p+q} \\
\iff
p\ln a-p\ln p +q\ln b-q\ln q\le (p+q)(\ln (a+b)-\ln (p+q)) \\
\iff
p(\ln a-\ln (a+b)+\ln(p+q)-\ln p)+q(\ln b-\ln (a+b)+\ln (p+q)-\ln q)\le 0 \\ \\
\xLeftrightarrow{\text{let } x=\dfrac{p}{p+q} ,y=\dfrac{a}{a+b} }
\\
x(\ln y-\ln x)+(1-x)(\ln (1-y)-\ln (1-x))\le 0 \\
f(y)=x\ln y+(1-x)\ln (1-y) \\
f'(y)=\dfrac{x}{y}-\dfrac{1-x}{1-y}=\dfrac{x-y}{y(1-y)} \\
\implies \operatorname{sign} f'(y)=-\operatorname{sign} (y-x),f(x) \text{ is a maximum } \\
\implies f(y)\le f(y)=0 \\
\\
\text{Q.E.D}
\end{gathered} ( p a ) p ( q b ) q ≤ ( p + q a + b ) p + q ⟺ p ln a − p ln p + q ln b − q ln q ≤ ( p + q ) ( ln ( a + b ) − ln ( p + q )) ⟺ p ( ln a − ln ( a + b ) + ln ( p + q ) − ln p ) + q ( ln b − ln ( a + b ) + ln ( p + q ) − ln q ) ≤ 0 let x = p + q p , y = a + b a x ( ln y − ln x ) + ( 1 − x ) ( ln ( 1 − y ) − ln ( 1 − x )) ≤ 0 f ( y ) = x ln y + ( 1 − x ) ln ( 1 − y ) f ′ ( y ) = y x − 1 − y 1 − x = y ( 1 − y ) x − y ⟹ sign f ′ ( y ) = − sign ( y − x ) , f ( x ) is a maximum ⟹ f ( y ) ≤ f ( y ) = 0 Q.E.D
T4
λ i > 0 , x i > 0 , ∑ i = 1 n λ i = 1 ⟹ ∏ i = 1 n x i λ i ≤ ∑ i = 1 n λ i x i \begin{gathered}
\lambda_i>0,x_i>0,\sum _{i = 1} ^{n} \lambda_i=1 \\
\implies \prod _{i = 1} ^{n} x_i^{\lambda_i}\le \sum _{i = 1} ^{n} \lambda_i x_i
\end{gathered} λ i > 0 , x i > 0 , i = 1 ∑ n λ i = 1 ⟹ i = 1 ∏ n x i λ i ≤ i = 1 ∑ n λ i x i
ln ( ∑ i = 1 n λ i x i ) ≥ ∑ i = 1 n λ i ln ( x i ) = ln ∏ i = 1 n x i λ i Q.E.D \begin{gathered}
\ln(\sum _{i = 1} ^{n} \lambda_ix_i)\ge \sum _{i = 1} ^{n} \lambda_i\ln(x_i)=\ln \prod _{i = 1} ^{n} x_i^{\lambda_i}\\
\text{Q.E.D}
\end{gathered} ln ( i = 1 ∑ n λ i x i ) ≥ i = 1 ∑ n λ i ln ( x i ) = ln i = 1 ∏ n x i λ i Q.E.D
T5
f ( x ) is convex in [ a , b ] , ∃ c ∈ ( a , b ) : f ( a ) = f ( c ) = f ( b ) ⟹ ∀ x ∈ [ a , b ] , f ( x ) = f ( a ) \begin{gathered}
f(x) \text{ is convex in } [a,b],\exists c\in (a,b):f(a)=f(c)=f(b) \\
\implies \forall x\in [a,b],f(x)=f(a)
\end{gathered} f ( x ) is convex in [ a , b ] , ∃ c ∈ ( a , b ) : f ( a ) = f ( c ) = f ( b ) ⟹ ∀ x ∈ [ a , b ] , f ( x ) = f ( a )
∀ x < c , { f ( x ) − f ( c ) x − c ≤ f ( c ) − f ( b ) c − b = 0 f ( x ) − f ( c ) x − c ≥ f ( c ) − f ( a ) c − a = 0 ⟹ f ( x ) = f ( c ) same for ∀ x > c , f ( x ) = c ⟹ ∀ x ∈ [ a , b ] , f ( x ) = f ( a ) \begin{gathered}
\forall x<c, \\
\begin{cases}
\dfrac{f(x)-f(c)}{x-c} \le \dfrac{f(c)-f(b)}{c-b} =0 \\
\dfrac{f(x)-f(c)}{x-c} \ge \dfrac{f(c)-f(a)}{c-a} =0
\end{cases} \\
\implies f(x)=f(c) \\
\text{same for } \forall x>c,f(x)=c \\
\implies \forall x\in [a,b],f(x)=f(a)
\end{gathered} ∀ x < c , ⎩ ⎨ ⎧ x − c f ( x ) − f ( c ) ≤ c − b f ( c ) − f ( b ) = 0 x − c f ( x ) − f ( c ) ≥ c − a f ( c ) − f ( a ) = 0 ⟹ f ( x ) = f ( c ) same for ∀ x > c , f ( x ) = c ⟹ ∀ x ∈ [ a , b ] , f ( x ) = f ( a )
T6
{ a < b < c < d f ( x ) is convex in [ a , c ] and [ b , d ] ⟹ f ( x ) is convex in [a,d] \begin{gathered}
\begin{cases}
a<b<c<d \\
f(x) \text{ is convex in } [a,c] \text{ and } [b,d] \\
\end{cases} \\
\implies f(x) \text{ is convex in [a,d]}
\end{gathered} { a < b < c < d f ( x ) is convex in [ a , c ] and [ b , d ] ⟹ f ( x ) is convex in [a,d]
∀ x 1 , x 2 ∈ [ a , d ] if x 1 , x 2 ∈ [ a , c ] f ( λ x 1 + ( 1 − λ ) x 2 ) ≤ λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) same for x 1 , x 2 ∈ [ b , d ] else x 1 ∈ [ a , b ) , x 2 ∈ ( c , d ] let m = λ x 1 + ( 1 − λ ) x 2 if m ∈ [ a , c ] f ( x 1 ) − f ( c ) x 1 − c ≤ f ( x 1 ) − f ( x 2 ) x 1 − x 2 ⟹ f ( m ) ≤ μ f ( x 1 ) + ( 1 − μ ) f ( c ) = f ( x 1 ) + f ( x 1 ) − f ( c ) x 1 − c ( m − x 1 ) ≤ f ( x 1 ) + f ( x 1 ) − f ( x 2 ) x 1 − x 2 ( m − x 1 ) = λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) same for m ∈ [ c , d ] ∴ f is convex in [ a , d ] \begin{gathered}
\forall x_1,x_2 \in [a,d] \\
\text{if } x_1,x_2\in [a,c] \\
f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2) \\
\text{same for } x_1,x_2\in[b,d] \\
\text{else } x_1\in [a,b),x_2\in (c,d] \\
\text{let } m=\lambda x_1+(1-\lambda)x_2 \\
\text{if } m \in [a,c] \\
\dfrac{f(x_1)-f(c)}{x_1-c} \le \dfrac{f(x_1)-f(x_2)}{x_1-x_2} \\
\implies
f(m)\le \mu f(x_1)+(1-\mu)f(c) \\
=f(x_1)+\dfrac{f(x_1)-f(c)}{x_1-c} (m-x_1) \\
\le f(x_1)+\dfrac{f(x_1)-f(x_2)}{x_1-x_2} (m-x_1) \\
=\lambda f(x_1)+(1-\lambda)f(x_2) \\
\text{same for } m\in[c,d] \\
\therefore f \text{ is convex in } [a,d]
\end{gathered} ∀ x 1 , x 2 ∈ [ a , d ] if x 1 , x 2 ∈ [ a , c ] f ( λ x 1 + ( 1 − λ ) x 2 ) ≤ λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) same for x 1 , x 2 ∈ [ b , d ] else x 1 ∈ [ a , b ) , x 2 ∈ ( c , d ] let m = λ x 1 + ( 1 − λ ) x 2 if m ∈ [ a , c ] x 1 − c f ( x 1 ) − f ( c ) ≤ x 1 − x 2 f ( x 1 ) − f ( x 2 ) ⟹ f ( m ) ≤ μ f ( x 1 ) + ( 1 − μ ) f ( c ) = f ( x 1 ) + x 1 − c f ( x 1 ) − f ( c ) ( m − x 1 ) ≤ f ( x 1 ) + x 1 − x 2 f ( x 1 ) − f ( x 2 ) ( m − x 1 ) = λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) same for m ∈ [ c , d ] ∴ f is convex in [ a , d ]
T7
f ( x ) is convex in I ⟺ ∀ c ∈ I , ∃ a , f ( x ) ≥ a ( x − c ) + f ( c ) \begin{gathered}
f(x) \text{ is convex in } I \\
\iff \forall c\in I,\exists a,f(x)\ge a(x-c)+f(c)
\end{gathered} f ( x ) is convex in I ⟺ ∀ c ∈ I , ∃ a , f ( x ) ≥ a ( x − c ) + f ( c )
反向:
∀ x 1 < x 2 , λ ∈ ( 0 , 1 ) let x 3 = λ x 1 + ( 1 − λ ) x 2 , k = f ( x 1 ) − f ( x 2 ) x 1 − x 2 ∃ a , s . t . { f ( x 1 ) ≥ a ( x 1 − x 3 ) + f ( x 3 ) f ( x 2 ) ≥ a ( x 2 − x 3 ) + f ( x 3 ) ⟹ { a ≤ f ( x 1 ) − f ( x 3 ) x 1 − x 3 a ≥ f ( x 2 ) − f ( x 3 ) x 2 − x 3 ⟹ f ( x 1 ) − f ( x 3 ) x 1 − x 3 ≥ f ( x 2 ) − f ( x 3 ) x 2 − x 3 ⟺ f ( x 3 ) ≤ λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) \begin{gathered}
\forall x_1<x_2,\lambda \in (0,1) \\
\text{let } x_3=\lambda x_1+(1-\lambda)x_2,k=\dfrac{f(x_1)-f(x_2)}{x_1-x_2} \\
\exists a, \\ s.t.\\
\begin{cases}
f(x_1)\ge a(x_1-x_3)+f(x_3) \\
f(x_2)\ge a(x_2-x_3)+f(x_3) \\
\end{cases} \\
\implies
\begin{cases}
a\le \dfrac{f(x_1)-f(x_3)}{x_1-x_3} \\
a\ge \dfrac{f(x_2)-f(x_3)}{x_2-x_3}
\end{cases} \\
\implies \dfrac{f(x_1)-f(x_3)}{x_1-x_3} \ge \dfrac{f(x_2)-f(x_3)}{x_2-x_3} \\
\iff f(x_3)\le \lambda f(x_1)+(1-\lambda)f(x_2)
\end{gathered} ∀ x 1 < x 2 , λ ∈ ( 0 , 1 ) let x 3 = λ x 1 + ( 1 − λ ) x 2 , k = x 1 − x 2 f ( x 1 ) − f ( x 2 ) ∃ a , s . t . { f ( x 1 ) ≥ a ( x 1 − x 3 ) + f ( x 3 ) f ( x 2 ) ≥ a ( x 2 − x 3 ) + f ( x 3 ) ⟹ ⎩ ⎨ ⎧ a ≤ x 1 − x 3 f ( x 1 ) − f ( x 3 ) a ≥ x 2 − x 3 f ( x 2 ) − f ( x 3 ) ⟹ x 1 − x 3 f ( x 1 ) − f ( x 3 ) ≥ x 2 − x 3 f ( x 2 ) − f ( x 3 ) ⟺ f ( x 3 ) ≤ λ f ( x 1 ) + ( 1 − λ ) f ( x 2 ) 正向:
let S = { f ( x ) − f ( c ) x − c ∣ x < c } T = { f ( x ) − f ( c ) x − c ∣ x > c } ∀ s ∈ S , t ∈ T : s < t ⟹ sup S ≤ inf T let a ∈ [ sup S , inf T ] ⟹ { ∀ x < c , f ( x ) − f ( c ) x − c < a ∀ x > c , f ( x ) − f ( c ) x − c > a ⟹ ∀ x , f ( x ) > a ( x − c ) \begin{gathered}
\text{let } S=\{ \dfrac{f(x)-f(c)}{x-c} \vert x<c \} \\
T=\{ \dfrac{f(x)-f(c)}{x-c} \vert x>c \} \\
\forall s\in S,t\in T:s<t \\
\implies \sup S\le \inf T \\
\text{let } a\in [\sup S,\inf T] \\
\implies \begin{cases}
\forall x<c,\dfrac{f(x)-f(c)}{x-c} <a \\
\forall x>c,\dfrac{f(x)-f(c)}{x-c} >a
\end{cases} \\
\implies \forall x,f(x)>a(x-c)
\end{gathered} let S = { x − c f ( x ) − f ( c ) ∣ x < c } T = { x − c f ( x ) − f ( c ) ∣ x > c } ∀ s ∈ S , t ∈ T : s < t ⟹ sup S ≤ inf T let a ∈ [ sup S , inf T ] ⟹ ⎩ ⎨ ⎧ ∀ x < c , x − c f ( x ) − f ( c ) < a ∀ x > c , x − c f ( x ) − f ( c ) > a ⟹ ∀ x , f ( x ) > a ( x − c )
[think] 就是直接构造a a a
T8
f ( x ) is convex in [ a , b ] ⟹ ∀ x ∈ [ a , b ] , f ( x ) ≤ max ( f ( a ) , f ( b ) ) \begin{gathered}
f(x) \text{ is convex in } [a,b] \\
\implies \forall x\in[a,b],f(x)\le \max(f(a),f(b))
\end{gathered} f ( x ) is convex in [ a , b ] ⟹ ∀ x ∈ [ a , b ] , f ( x ) ≤ max ( f ( a ) , f ( b ))
let m = max ( f ( a ) , f ( b ) ) λ = x − a b − a ⟹ f ( x ) ≤ λ f ( a ) + ( 1 − λ ) f ( b ) ≤ λ m + ( 1 − λ ) m = m Q.E.D \begin{gathered}
\text{let } m=\max(f(a),f(b)) \\
\lambda=\dfrac{x-a}{b-a} \\
\implies
f(x)\le \lambda f(a)+(1-\lambda)f(b) \\
\le \lambda m + (1-\lambda) m \\
=m \\
\text{Q.E.D}
\end{gathered} let m = max ( f ( a ) , f ( b )) λ = b − a x − a ⟹ f ( x ) ≤ λ f ( a ) + ( 1 − λ ) f ( b ) ≤ λm + ( 1 − λ ) m = m Q.E.D
T9
f ( x ) is convex in [ a , b ] ⟹ f ( x ) is bounded in [ a , b ] \begin{gathered}
f(x) \text{ is convex in } [a,b] \\
\implies f(x) \text{ is bounded in } [a,b]
\end{gathered} f ( x ) is convex in [ a , b ] ⟹ f ( x ) is bounded in [ a , b ]
According to T8, f ( x ) ≤ M 1 = max ( f ( a ) , f ( b ) ) let c ∈ ( a , b ) ∀ x < c , f ( c ) − f ( x ) c − x ≤ f ( b ) − f ( c ) b − c ⟹ f ( x ) ≥ f ( c ) − f ( b ) − f ( c ) b − c ( c − x ) ≥ f ( c ) − f ( b ) − f ( c ) b − c ( c − a ) = M 2 same for x > c , f ( x ) ≥ f ( c ) − f ( c ) − f ( a ) c − a ( b − c ) = M 3 ⟹ ∣ f ( x ) ∣ ≤ max ( ∣ M 1 ∣ , ∣ M 2 ∣ , ∣ M 3 ∣ ) = M \begin{gathered}
\text{According to T8,} f(x)\le M_1=\max (f(a),f(b)) \\ \text{let } c\in (a,b) \\
\forall x<c,\dfrac{f(c)-f(x)}{c-x} \le \dfrac{f(b)-f(c)}{b-c} \\
\implies f(x)\ge f(c)-\dfrac{f(b)-f(c)}{b-c} (c-x) \\
\ge f(c)-\dfrac{f(b)-f(c)}{b-c} (c-a)=M_2 \\
\text{same for } x>c,f(x)\ge f(c)-\dfrac{f(c)-f(a)}{c-a} (b-c)=M_3 \\
\implies \vert f(x) \vert \le \max (\vert M_1 \vert ,\vert M_2 \vert ,\vert M_3 \vert) =M
\end{gathered} According to T8, f ( x ) ≤ M 1 = max ( f ( a ) , f ( b )) let c ∈ ( a , b ) ∀ x < c , c − x f ( c ) − f ( x ) ≤ b − c f ( b ) − f ( c ) ⟹ f ( x ) ≥ f ( c ) − b − c f ( b ) − f ( c ) ( c − x ) ≥ f ( c ) − b − c f ( b ) − f ( c ) ( c − a ) = M 2 same for x > c , f ( x ) ≥ f ( c ) − c − a f ( c ) − f ( a ) ( b − c ) = M 3 ⟹ ∣ f ( x ) ∣ ≤ max ( ∣ M 1 ∣ , ∣ M 2 ∣ , ∣ M 3 ∣ ) = M
Class 2
T1
lim x → π 2 ln sin ( x ) ( π − 2 x ) 2 \begin{gathered}
\lim_{x \to \frac{\pi}{2} } \dfrac{\ln \sin(x)}{(\pi-2x)^2}
\end{gathered} x → 2 π lim ( π − 2 x ) 2 ln sin ( x )
lim x → π 2 ln sin ( x ) ( π − 2 x ) 2 = lim x → π 2 cos x sin x 4 ( 2 x − π ) = lim x → π 2 cos x 4 ( 2 x − π ) = lim x → π 2 − sin x 8 = − 1 8 \begin{gathered}
\lim_{x \to \frac{\pi}{2} } \dfrac{\ln \sin(x)}{(\pi-2x)^2} \\
=\lim_{x \to \frac{\pi}{2} } \dfrac{\dfrac{\cos x}{\sin x} }{4(2x-\pi)} \\
=\lim_{x \to \frac{\pi}{2} } \dfrac{\cos x}{4(2x-\pi)} \\
=\lim_{x \to \frac{\pi}{2} } \dfrac{-\sin x}{8} \\
=-\dfrac{1}{8}
\end{gathered} x → 2 π lim ( π − 2 x ) 2 ln sin ( x ) = x → 2 π lim 4 ( 2 x − π ) sin x cos x = x → 2 π lim 4 ( 2 x − π ) cos x = x → 2 π lim 8 − sin x = − 8 1
T2
lim x → 1 − ln x ln ( 1 − x ) \begin{gathered}
\lim_{x \to 1^-} \ln x\ln(1-x)
\end{gathered} x → 1 − lim ln x ln ( 1 − x )
lim x → 1 − ln x ln ( 1 − x ) = lim x → 1 − ln ( 1 + ( x − 1 ) ) ln ( 1 − x ) = lim x → 1 − ( x − 1 ) ln ( 1 − x ) = lim x → 1 − ln ( 1 − x ) 1 x − 1 = lim x → 1 − − 1 1 − x − 1 ( x − 1 ) 2 = 0 \begin{gathered}
\lim_{x \to 1^-} \ln x\ln(1-x) \\
=\lim_{x \to 1^-} \ln(1+(x-1))\ln(1-x) \\
=\lim_{x \to 1^-} (x-1)\ln(1-x) \\
=\lim_{x \to 1^-} \dfrac{\ln(1-x)}{\dfrac{1}{x-1} } \\
=\lim_{x \to 1^-} \dfrac{-\dfrac{1}{1-x} }{-\dfrac{1}{(x-1)^2} } =0
\end{gathered} x → 1 − lim ln x ln ( 1 − x ) = x → 1 − lim ln ( 1 + ( x − 1 )) ln ( 1 − x ) = x → 1 − lim ( x − 1 ) ln ( 1 − x ) = x → 1 − lim x − 1 1 ln ( 1 − x ) = x → 1 − lim − ( x − 1 ) 2 1 − 1 − x 1 = 0
T3
lim x → ∞ x [ ( 1 + 1 x ) x − e ] \begin{gathered}
\lim_{x \to \infty} x[(1+\dfrac{1}{x} )^x-e]
\end{gathered} x → ∞ lim x [( 1 + x 1 ) x − e ]
lim x → ∞ x [ ( 1 + 1 x ) x − e ] = lim x → ∞ ( 1 + 1 x ) x − e 1 x = lim x → 0 e ln ( x + 1 ) x − e x = lim x → 0 e ln ( 1 + x ) x ⋅ x 1 + x − ln ( 1 + x ) x 2 = lim x → 0 e 1 ( 1 + x ) 2 − 1 1 + x 2 x = − 1 2 e \begin{gathered}
\lim_{x \to \infty} x[(1+\dfrac{1}{x} )^x-e] \\
=\lim_{x \to \infty} \dfrac{(1+\dfrac{1}{x} )^x-e}{\dfrac{1}{x} } \\
=\lim_{x \to 0} \dfrac{e^{\frac{\ln (x+1)}x}-e}{x} \\
=\lim_{x \to 0} e^{\frac{\ln (1+x)}{x} } \cdot \dfrac{\frac{x}{1+x}-\ln(1+x)}{x^2} \\
=\lim_{x \to 0} e \dfrac{\dfrac{1}{(1+x)^2} -\dfrac{1}{1+x} }{2x} \\
=-\dfrac{1}{2} e
\end{gathered} x → ∞ lim x [( 1 + x 1 ) x − e ] = x → ∞ lim x 1 ( 1 + x 1 ) x − e = x → 0 lim x e x l n ( x + 1 ) − e = x → 0 lim e x l n ( 1 + x ) ⋅ x 2 1 + x x − ln ( 1 + x ) = x → 0 lim e 2 x ( 1 + x ) 2 1 − 1 + x 1 = − 2 1 e
T4
lim x → 0 ( ( 1 + x ) 1 x e ) 1 x \begin{gathered}
\lim_{x \to 0} {\left( \dfrac{(1+x)^\frac1x}{e} \right)}^{\frac{1}{x}}
\end{gathered} x → 0 lim ( e ( 1 + x ) x 1 ) x 1
lim x → 0 exp 1 x ( 1 x ln ( 1 + x ) − 1 ) = lim x → 0 exp ln ( 1 + x ) − x x 2 = exp lim x → 0 ln ( 1 + x ) − x x 2 = exp lim x → 0 1 1 + x − 1 2 x = e − 1 2 \begin{gathered}
\lim_{x \to 0} \exp \dfrac{1}{x} {\left( \dfrac{1}{x} \ln(1+x)-1 \right)} \\
=\lim_{x \to 0} \exp \dfrac{\ln(1+x)-x}{x^2} \\
=\exp \lim_{x\to 0}\dfrac{\ln(1+x)-x }{x^2} \\ \\
=\exp \lim_{x\to 0}\dfrac{\dfrac{1}{1+x} -1}{2x}
=e^{-\frac12}
\end{gathered} x → 0 lim exp x 1 ( x 1 ln ( 1 + x ) − 1 ) = x → 0 lim exp x 2 ln ( 1 + x ) − x = exp x → 0 lim x 2 ln ( 1 + x ) − x = exp x → 0 lim 2 x 1 + x 1 − 1 = e − 2 1
T5
{ f ( x ) ∈ C 2 ( a , + ∞ ) lim x → + ∞ ( f ( x ) + 2 f ′ ( x ) + f ′ ′ ( x ) ) = l ⟹ { lim x → + ∞ f ( x ) = l lim x → + ∞ f ′ ( x ) = lim x → + ∞ f ′ ′ ( x ) = 0 \begin{gathered}
\begin{cases}
f(x)\in C^2(a,+\infty) \\
\lim_{x \to +\infty} (f(x)+2f'(x)+f''(x))=l
\end{cases} \\
\implies \begin{cases}
\lim_{x \to +\infty} f(x)=l \\
\lim_{x \to +\infty} f'(x)=\lim_{x \to +\infty} f''(x)=0
\end{cases}
\end{gathered} { f ( x ) ∈ C 2 ( a , + ∞ ) lim x → + ∞ ( f ( x ) + 2 f ′ ( x ) + f ′′ ( x )) = l ⟹ { lim x → + ∞ f ( x ) = l lim x → + ∞ f ′ ( x ) = lim x → + ∞ f ′′ ( x ) = 0
let F ( x ) = e x f ( x ) ⟹ { F ′ ( x ) = e x ( f ( x ) + f ′ ( x ) ) F ′ ′ ( x ) = e x ( f ( x ) + 2 f ′ ( x ) + f ′ ′ ( x ) ) ⟹ lim x → + ∞ F ′ ′ ( x ) e x = l lim x → + ∞ F ( x ) e x = lim x → + ∞ F ′ ( x ) e x = lim x → + ∞ F ′ ′ ( x ) e x = l ⟹ { lim x → + ∞ f ( x ) = l lim x → + ∞ f ′ ( x ) = lim x → + ∞ F ′ ( x ) e x − f ( x ) = 0 lim x → + ∞ f ′ ′ ( x ) = lim x → + ∞ F ′ ′ ( x ) e x − 2 f ′ ( x ) − f ( x ) = 0 \begin{gathered}
\text{let } F(x)=e^xf(x) \\
\implies
\begin{cases}
F'(x)=e^x(f(x)+f'(x)) \\
F''(x)=e^x (f(x)+2f'(x)+f''(x))
\end{cases} \\
\implies \lim_{x \to +\infty} \dfrac{F''(x)}{e^x}=l \\
\lim_{x \to +\infty} \dfrac{F(x)}{e^x} \\
=\lim_{x \to +\infty} \dfrac{F'(x)}{e^x} \\
=\lim_{x \to +\infty} \dfrac{F''(x)}{e^x} =l \\
\implies
\begin{cases}
\lim_{x \to +\infty} f(x)=l \\
\lim_{x \to +\infty} f'(x)=\lim_{x \to +\infty} \dfrac{F'(x)}{e^x} -f(x)=0 \\
\lim_{x \to +\infty} f''(x)=\lim_{x \to +\infty} \dfrac{F''(x)}{e^x} -2f'(x)-f(x)=0
\end{cases}
\end{gathered} let F ( x ) = e x f ( x ) ⟹ { F ′ ( x ) = e x ( f ( x ) + f ′ ( x )) F ′′ ( x ) = e x ( f ( x ) + 2 f ′ ( x ) + f ′′ ( x )) ⟹ x → + ∞ lim e x F ′′ ( x ) = l x → + ∞ lim e x F ( x ) = x → + ∞ lim e x F ′ ( x ) = x → + ∞ lim e x F ′′ ( x ) = l ⟹ ⎩ ⎨ ⎧ lim x → + ∞ f ( x ) = l lim x → + ∞ f ′ ( x ) = lim x → + ∞ e x F ′ ( x ) − f ( x ) = 0 lim x → + ∞ f ′′ ( x ) = lim x → + ∞ e x F ′′ ( x ) − 2 f ′ ( x ) − f ( x ) = 0
T6
∃ f ′ ′ ( x 0 ) , f ′ ( x 0 ) ≠ 0 calc lim x → x 0 ( 1 f ( x ) − f ( x 0 ) − 1 ( x − x 0 ) f ′ ( x 0 ) ) \begin{gathered}
\exists f''(x_0),f'(x_0)\ne 0 \\
\text{calc } \lim_{x \to x_0} {\left( \dfrac{1}{f(x)-f(x_0)} -\dfrac{1}{(x-x_0)f'(x_0)} \right)}
\end{gathered} ∃ f ′′ ( x 0 ) , f ′ ( x 0 ) = 0 calc x → x 0 lim ( f ( x ) − f ( x 0 ) 1 − ( x − x 0 ) f ′ ( x 0 ) 1 )
lim x → x 0 ( 1 f ( x ) − f ( x 0 ) − 1 ( x − x 0 ) f ′ ( x 0 ) ) = lim x → x 0 ( x − x 0 ) f ′ ( x 0 ) − ( f ( x ) − f ( x 0 ) ) ( f ( x ) − f ( x 0 ) ) ( x − x 0 ) f ′ ( x 0 ) = lim x → x 0 f ′ ( x 0 ) − f ′ ( x ) f ′ ( x 0 ) ( f ( x ) + x f ′ ( x ) − x 0 f ′ ( x ) − f ( x 0 ) ) = lim x → x 0 − f ′ ′ ( x ) f ′ ( x 0 ) ( 2 f ′ ( x ) + ( x − x 0 ) f ′ ′ ( x ) ) = − f ′ ′ ( x 0 ) 2 ( f ′ ( x 0 ) ) 2 \begin{gathered}
\lim_{x \to x_0} {\left( \dfrac{1}{f(x)-f(x_0)} -\dfrac{1}{(x-x_0)f'(x_0)} \right)} \\
=\lim_{x \to x_0} \dfrac{(x-x_0)f'(x_0)-(f(x)-f(x_0))}{(f(x)-f(x_0))(x-x_0)f'(x_0)} \\
=\lim_{x \to x_0} \dfrac{f'(x_0)-f'(x)}{f'(x_0)(f(x)+xf'(x)-x_0f'(x)-f(x_0))} \\
=\lim_{x \to x_0} \dfrac{-f''(x)}{f'(x_0)(2f'(x)+(x-x_0)f''(x))} \\
=-\dfrac{f''(x_0)}{2(f'(x_0))^2}
\end{gathered} x → x 0 lim ( f ( x ) − f ( x 0 ) 1 − ( x − x 0 ) f ′ ( x 0 ) 1 ) = x → x 0 lim ( f ( x ) − f ( x 0 )) ( x − x 0 ) f ′ ( x 0 ) ( x − x 0 ) f ′ ( x 0 ) − ( f ( x ) − f ( x 0 )) = x → x 0 lim f ′ ( x 0 ) ( f ( x ) + x f ′ ( x ) − x 0 f ′ ( x ) − f ( x 0 )) f ′ ( x 0 ) − f ′ ( x ) = x → x 0 lim f ′ ( x 0 ) ( 2 f ′ ( x ) + ( x − x 0 ) f ′′ ( x )) − f ′′ ( x ) = − 2 ( f ′ ( x 0 ) ) 2 f ′′ ( x 0 )
T7
∣ x ∣ < 1 , arcsin x = x 1 − θ 2 x 2 , θ ∈ ( 0 , 1 ) calc lim x → 0 θ \begin{gathered}
\vert x \vert <1,\arcsin x=\dfrac{x}{\sqrt{1-\theta^2x^2}} ,\theta\in (0,1) \\
\text{calc } \lim_{x \to 0}\theta
\end{gathered} ∣ x ∣ < 1 , arcsin x = 1 − θ 2 x 2 x , θ ∈ ( 0 , 1 ) calc x → 0 lim θ
θ 2 = 1 x 2 − 1 arcsin 2 ( x ) = arcsin 2 ( x ) − x 2 x 2 arcsin 2 ( x ) let t = arcsin ( x ) , x = sin t ⟹ θ 2 = t 2 − sin 2 t t 2 sin 2 t lim x → 0 θ 2 = lim t → 0 t 2 − sin 2 t t 2 sin 2 t = lim t → 0 t 2 − sin 2 t t 4 = lim t → 0 2 t − sin 2 t 4 t 3 = lim t → 0 1 − cos 2 t 6 t 2 = lim t → 0 t 2 3 t 2 = 1 3 ⟹ lim x → 0 θ = 3 3 \begin{gathered}
\theta^2=\dfrac{1}{x^2} -\dfrac{1}{\arcsin^2(x)} \\
=\dfrac{\arcsin^2(x)-x^2}{x^2\arcsin^2(x)} \\
\text{let } t=\arcsin(x),x=\sin t \\
\implies \theta^2=\dfrac{t^2-\sin^2 t}{t^2\sin^2 t} \\
\lim_{x \to 0} \theta^2 \\
=\lim_{t \to 0} \dfrac{t^2-\sin^2t}{t^2\sin^2t} \\
=\lim_{t \to 0} \dfrac{t^2-\sin^2 t}{t^4} \\
=\lim_{t \to 0} \dfrac{2t-\sin 2t}{4t^3} \\
=\lim_{t \to 0} \dfrac{1-\cos2t}{6t^2} \\
=\lim_{t \to 0} \dfrac{t^2}{3t^2} \\
=\dfrac{1}{3} \\
\implies \lim_{x \to 0} \theta = \dfrac{\sqrt 3}{3}
\end{gathered} θ 2 = x 2 1 − arcsin 2 ( x ) 1 = x 2 arcsin 2 ( x ) arcsin 2 ( x ) − x 2 let t = arcsin ( x ) , x = sin t ⟹ θ 2 = t 2 sin 2 t t 2 − sin 2 t x → 0 lim θ 2 = t → 0 lim t 2 sin 2 t t 2 − sin 2 t = t → 0 lim t 4 t 2 − sin 2 t = t → 0 lim 4 t 3 2 t − sin 2 t = t → 0 lim 6 t 2 1 − cos 2 t = t → 0 lim 3 t 2 t 2 = 3 1 ⟹ x → 0 lim θ = 3 3
Class 3
T1
T2
calc ∣ S ∣ ∣ S ∣ = { x ∣ ln x − x e = k , x ∈ ( 0 , + ∞ ) } \begin{gathered}
\text{calc } \vert S \vert \\
\vert S \vert = \{ x \vert \ln x-\dfrac{x}{e} =k,x\in(0,+\infty) \}
\end{gathered} calc ∣ S ∣ ∣ S ∣ = { x ∣ ln x − e x = k , x ∈ ( 0 , + ∞ )}
let f ( x ) = ln x − x e f ′ ( x ) = 1 x − 1 e ⟹ { x < e ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > e ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing f ( e ) = 0 is maximum of f { lim x → + ∞ f ( x ) = − ∞ lim x → 0 + f ( x ) = − ∞ ⟹ { ∣ S ∣ = 0 , k > 0 ∣ S ∣ = 1 , k = 0 ∣ S ∣ = 2 , k < 0 \begin{gathered}
\text{let } f(x)=\ln x-\dfrac{x}{e} \\
f'(x)=\dfrac{1}{x} -\dfrac{1}{e} \\
\implies \begin{cases}
x<e \implies f'(x)>0,f(x) \text{ is increasing} \\
x>e \implies f'(x)<0,f(x) \text{ is decreasing} \\
f(e)=0 \text{ is maximum of } f \\
\end{cases} \\
\begin{cases}
\lim_{x \to +\infty} f(x)=-\infty \\
\lim_{x \to 0^+} f(x)=-\infty \\
\end{cases} \\
\implies \begin{cases}
\vert S \vert =0,k>0 \\
\vert S \vert =1,k=0 \\
\vert S \vert =2,k<0
\end{cases}
\end{gathered} let f ( x ) = ln x − e x f ′ ( x ) = x 1 − e 1 ⟹ ⎩ ⎨ ⎧ x < e ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > e ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing f ( e ) = 0 is maximum of f { lim x → + ∞ f ( x ) = − ∞ lim x → 0 + f ( x ) = − ∞ ⟹ ⎩ ⎨ ⎧ ∣ S ∣ = 0 , k > 0 ∣ S ∣ = 1 , k = 0 ∣ S ∣ = 2 , k < 0
T3
x > 0 ⟹ ∃ ! x 0 , k x + 1 x 0 2 = 1 solve k \begin{gathered}
x>0 \implies \exists !x_0,kx+\dfrac{1}{x_0^2} =1 \\
\text{solve } k
\end{gathered} x > 0 ⟹ ∃ ! x 0 , k x + x 0 2 1 = 1 solve k
let f ( x ) = x 2 − 1 x 3 f ′ ( x ) = 3 − x 2 x 4 ⟹ { x < 3 ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > 3 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing f ( 3 ) = 2 3 9 is maximum { lim x → 0 + f ( x ) = − ∞ lim x → + ∞ f ( x ) = 0 ⟹ k ∈ { 2 3 9 } ∪ ( − ∞ , 0 ] \begin{gathered}
\text{let } f(x)=\dfrac{x^2-1}{x^3} \\
f'(x)=\dfrac{3-x^2}{x^4} \\
\implies \begin{cases}
x<\sqrt 3 \implies f'(x)>0,f(x) \text{ is increasing} \\
x>\sqrt 3 \implies f'(x)<0,f(x) \text{ is decreasing} \\
f(\sqrt 3)=\dfrac{2\sqrt 3}{9} \text{ is maximum}
\end{cases} \\
\begin{cases}
\lim_{x \to 0^+} f(x)=-\infty \\
\lim_{x \to +\infty} f(x)=0
\end{cases} \\
\implies k \in \{ \dfrac{2\sqrt 3}{9} \} \cup (-\infty,0]
\end{gathered} let f ( x ) = x 3 x 2 − 1 f ′ ( x ) = x 4 3 − x 2 ⟹ ⎩ ⎨ ⎧ x < 3 ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > 3 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing f ( 3 ) = 9 2 3 is maximum { lim x → 0 + f ( x ) = − ∞ lim x → + ∞ f ( x ) = 0 ⟹ k ∈ { 9 2 3 } ∪ ( − ∞ , 0 ]
