Differentiation using l´Hopital

For the first one we have

$$(1-\sin x)^{\cos x}= (1-\sin x)^{\cos x}\frac{(1+\sin x)^{\cos x}}{(1+\sin x)^{\cos x}}=\frac{(\cos^2 x)^{\cos x}}{(1+\sin x)^{\cos x}}\to \frac 11=1$$

indeed by standard limits

  • $\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2 x)^{\cos x}=\lim_{t\to 0}(t^2)^t=1$

  • $\lim_{x\rightarrow\frac{\pi}{2}} (1+\sin x)^{\cos x}=2^0=1$

For the second one

$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan x)^{\tan(2x)}=\lim_{x\rightarrow\frac{\pi}{4}} \left[(1+(\tan x-1))^\frac{1}{\tan x-1}\right]^{\tan(2x)(\tan x-1)}\to e^{-1}=\frac1e$$

indeed by standard limits

  • $\lim_{x\rightarrow\frac{\pi}{4}} (1+(\tan x-1))^\frac{1}{\tan x-1}\to e$

  • $\lim_{x\rightarrow \frac{\pi}{4}} \tan(2x)(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{2\tan x}{1-\tan^2 x}(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{-2\tan x}{1+\tan x}=-1$


$$\begin{align} \lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\ln(1-\sin(x))^{\cos(x)}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\cos(x)\ln(1-\sin(x))\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\ln(1-\sin(x))}{1/\cos(x)}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\frac{-\cos(x)}{1-\sin(x)}}{\frac{\sin(x)}{\cos^2(x)}}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos^3(x)}{(\sin(x)-1)\sin(x)}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos^3(x)}{\sin(x)-1}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{-3\cos^2(x)\sin(x)}{\cos(x)}\bigg)\\ &=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}-3\cos(x)\sin(x)\bigg)\\ &=e^0\\ &=1 \end{align}$$ I can tell you the answer for the second limit is $1/e$, but the technique is pretty much the same.


Without using L'Hospital: $$\begin{align}1) \ \lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot 1=\\ &=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot \lim_{x\rightarrow\frac{\pi}{2}} (1+\sin(x))^{\cos(x)}=\\ &=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin^2(x))^{\cos(x)}=\\ &=\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2(x))^{\cos(x)}\stackrel{(1)}=\\ &=1.\\ 2) \ \lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}&=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{\frac{2\tan(x)}{(1-\tan x)(1+\tan x)}}=\\ &=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{-\frac{1}{\tan x-1}}\stackrel{(2)}=\\ &=e^{-1}.\end{align}$$ Note: $$\begin{align}\lim_\limits{x\to 0} x^x&=1 \quad (1)\\ \lim_\limits{x\to 0} (1+x)^{1/x}&=e \quad (2) \end{align}$$

Tags:

Limits

Calculus

Algebra Precalculus

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