# Thread: Limit without using L'Hospital Rule

1. ## Limit without using L'Hospital Rule

I've already solved this porblem using L'Hospital rule, and now I'm required to solve it without using this rule, rather using the given condition.

Find:

Lim (sinx)^tanx
x-->pi/2

Given that: Lim (1+1/x)^x= e
x-->infinity

The answer is, as you might have guessed, 1. But, can anyone do it absolutely without using L'Hopital rule? I tried to do, but it was only elegant solution, so that's worthless.

2. a) considering the identity...

$(\sin x)^{\tan x} = e^{\tan x\cdot \ln \sin x}$ (1)

... and the fact that is...

$\ln \sin x = \frac{1}{2}\cdot \ln (1- \cos^{2} x) = - \frac{1}{2}\cdot (\cos^{2} x + \frac{\cos ^{4} x}{2} + \dots)$ (2)

... so that is...

$\lim_{x \rightarrow \frac {\pi}{2}} \frac{\sin x\cdot \ln \sin x}{\cos x} = 0$ (3)

... we conclude that is...

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

b) considering that for $x>1$ is...

$\ln \{(1+\frac{1}{x})^{x}\} = x \cdot \ln (1+\frac{1}{x}) = x \cdot (\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - \dots)$ (5)

... we conclude that is...

$\lim_{ x \rightarrow \infty} (1+\frac{1}{x})^{x} = e$ (6)

Kind regards

$\chi$ $\sigma$

3. Hey, your no. (2) is completely wrong. You get it in the form (0/0) not negative infinity. So the solution is just elegant. I just need to find the limit given that condition (meaning using that condition, instead of L'Hospital rule). So... if anyone can please help me.

Originally Posted by chisigma
a) considering the identity...

$(\sin x)^{\tan x} = e^{\tan x\cdot \ln \sin x}$ (1)

... and the fact that is...

$\lim_{x \rightarrow \frac {\pi}{2}} \frac{\sin x\cdot \ln \sin x}{\cos x} = - \infty$ (2)

... we conclude that is...

$\lim_{x \rightarrow \frac {\pi}{2}} (\sin x)^{\tan x} =0$ (3)

b) considering that for $x>1$ is...

$\ln \{(1+\frac{1}{x})^{x}\} = x \cdot \ln (1+\frac{1}{x}) = x \cdot (\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - \dots)$ (4)

... we conclude that is...

$\lim_{ x \rightarrow \infty} (1+\frac{1}{x})^{x} = e$ (5)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by concern
I've already solved this porblem using L'Hospital rule, and now I'm required to solve it without using this rule, rather using the given condition.

Find:

Lim (sinx)^tanx
x-->pi/2

Given that: Lim (1+1/x)^x= e
x-->infinity

The answer is, as you might have guessed, 1. But, can anyone do it absolutely without using L'Hopital rule? I tried to do, but it was only elegant solution, so that's worthless.
Problem: Compute $L=\lim_{x\to\frac{\pi}{2}}\sin(x)^{\tan(x)}$ without use of L'hopital's rule

Solution (1): Let $x=\frac{\pi}{2}-z$ so $L=\lim_{z\to0}\sin\left(\frac{\pi}{2}-z\right)^{\tan\left(\frac{\pi}{2}-z\right)}=\lim_{z\to0}\cos(z)^{\cot(z)}$. Finally we can see that $L=\lim_{z\to0}\cos(z)^{\cot(z)}\implies\ln\left(L\ right)=\lim_{z\to0}\cot(z)\ln\left(\cos(z)\right)$. Now in the neighborhood of zero $\cos(z)\stackrel{0}\sim 1-\frac{z^2}{2}$ so that $\ln\left(L\right)=\lim_{z\to0}\cot(z)\ln\left(1-\frac{z^2}{2}\right)$ and lastly noting that $\ln\left(1-\frac{z^2}{2}\right)\stackrel{0}\sim \frac{z^2}{2}$ we may conclude that $\ln\left(L\right)=\lim_{z\to0}\frac{z^2}{2}\cot(z) =\frac{1}{2}\lim_{z\to0}\frac{z^2}{\tan(z)}$. And once again appealing to asymptotic equivalence $\tan(z)\stackrel{0}\sim z$ so $\ln\left(L\right)=\frac{1}{2}\lim_{z\to0}\frac{z^2 }{z}=0\implies L=e^0=1$.

Solution (2): Let $\tan(x)=z$ so that $L=\lim_{z\to\infty}\sin\left(\text{arctan}(z)\righ t)^z=\lim_{z\to\infty}\left(\frac{z}{\sqrt{z^2+1}} \right)^z$. Letting $z=\frac{1}{\ell}$ turns this limit into $L=\lim_{\ell\to0}\left(\frac{\frac{1}{\ell}}{\sqrt {\frac{1}{\ell^2}+1}}\right)^{\frac{1}{\ell}}=\lim _{\ell\to0}\left(\frac{1}{\sqrt{\ell^2+1}}\right)^ \frac{1}{\ell}\implies\ln\left(L\right)=\frac{-1}{2}\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right) }{\ell}$. But $\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)}{\el l}=\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)-\ln\left(1+0^2\right)}{\ell-0}=\bigg[ \ln\left(1+\ell^2\right)\bigg]'\bigg|_{\ell=0}=\bigg[\frac{2\ell}{\ell^2+1}\bigg]\bigg|_{\ell=0}=0$ (why?). So that $\ln\left(L\right)=\frac{-1}{2}\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right) }{\ell}=\frac{-1}{2}\cdot0=0\implies L=e^0=1$

5. Thanks Drexel, this works perfectly fine.