Question on Limits

**h2osprey** · June 28th 2008, 08:40 AM

Find lim as x-->0+ for x^(tan x).

It seems to suggest 0^0 = 1, but when I apply L'Hopital's I get 0 instead. Help would be great

edit: sorry missed out the minus sign, should be 0 instead of infinity.

**Moo** · June 28th 2008, 08:46 AM

Hello,

Originally Posted by h2osprey

Find lim as x-->0+ for x^(tan x).

It seems to suggest 0^0 = 1, but when I apply L'Hopital's I get infinity instead. Help would be great

L'Hôpital's rule must be used only if you have $\frac{\infty}{\infty}$ or $\frac 00$ ...

The answer is indeed 1.

$x^{\tan x}=e^{\tan x \ln x}$

$\tan x \sim x$ when $x \to 0$

and $\lim_{x \to 0^+} x \ln x=0$

---> $\lim_{x \to 0^+} x^{\tan x}=\lim_{x \to 0^+} e^{\tan x \ln x}=\lim_{x \to 0^+} e^{x \ln x}=e^0=1$

**Mathstud28** · June 28th 2008, 08:50 AM

Originally Posted by h2osprey

Find lim as x-->0+ for x^(tan x).

It seems to suggest 0^0 = 1, but when I apply L'Hopital's I get 0 instead. Help would be great

edit: sorry missed out the minus sign, should be 0 instead of infinity.

$\color{red}0^0\ne{1}$

Let $\tan(x)=\frac{1}{\varphi}\Rightarrow{\arctan\left( \frac{1}{\varphi}\right)=x}$

So as $x\to{0}\Rightarrow\varphi\to\infty$

So we have

$\lim_{\varphi\to\infty}\arctan\left(\frac{1}{\varp hi}\right)^\frac{1}{\varphi}$

Now by the relation of the Root and Ratio test this is equivalent to

$\lim_{\varphi\to\infty}\frac{\arctan\left(\frac{1} {\varphi+1}\right)}{\arctan\left(\frac{1}{\varphi} \right)}\sim\lim_{\varphi\to\infty}\frac{\frac{1}{ \varphi+1}}{\frac{1}{\varphi}}=1$

$\therefore\quad\lim_{x\to{0}}x^{\tan(x)}=1$

**h2osprey** · June 28th 2008, 08:53 AM

Originally Posted by Moo

Hello,

L'Hôpital's rule must be used only if you have $\frac{\infty}{\infty}$ or $\frac 00$ ...

The answer is indeed 1.

$x^{\tan x}=e^{\tan x \ln x}$

$\tan x \sim x$ when $x \to 0$

and $\lim_{x \to 0^+} x \ln x=0$

---> $\lim_{x \to 0^+} x^{\tan x}=\lim_{x \to 0^+} e^{\tan x \ln x}=\lim_{x \to 0^+} e^{x \ln x}=e^0=1$

Thanks.
But how would you evaluate

$\lim_{x \to 0^+} x \ln x$

without L'Hôpital's?

**Krizalid** · June 28th 2008, 09:05 AM

$x^{\tan x}=\exp \left\{ \tan x\ln x \right\}=\exp \left\{ \frac{\tan x}{x}\cdot x\ln x \right\},$ now by takin' the limit we have fairly known expressions there.

**Mathstud28** · June 28th 2008, 09:08 AM

Originally Posted by h2osprey

Thanks.
But how would you evaluate lim x->0+ for xlnx without L'Hopital's?

$\lim_{x\to{0}}x\ln(x)$

There are many ways, but the easies would probably be

Let $\xi=\frac{1}{x}\Rightarrow\frac{1}{\xi}=x$

So as $x\to{0}\Rightarrow\xi\to\infty$

So we would have

$\lim_{\xi\to\infty}\frac{\ln\left(\frac{1}{\xi}\ri ght)}{\xi}=\lim_{\xi\to\infty}\frac{-\ln(\xi)}{\xi}$

Now we can say one of two things

$\ln(x)\prec{x}\quad\therefore\lim_{\xi\to\infty}\f rac{-\ln(\xi)}{\xi}=0$

Or

$\forall{\xi}>1\quad{0\geq\frac{\ln(\xi)}{\xi}}$

And

$\exists{N}\backepsilon\forall{\xi>N}\quad\frac{\ln (\xi)}{\xi}\leq\frac{1}{\sqrt{\xi}}$

This implies that

$\exists{N}\backepsilon\forall\xi>N\quad{0\leq\frac {\ln(\xi)}{\xi}\leq\frac{1}{\sqrt{\xi}}}$

Now remaining inequality

$\exists{N}\backepsilon\forall{\xi>N}\quad\lim_{\xi \to\infty}0\leq\lim_{\xi\to\infty}\frac{\ln(\xi)}{ \xi}\leq\lim_{\xi\to\infty}\frac{1}{\sqrt{\xi}}$

Now the left and right limits are obviously zero, so we have

$0\leq\lim_{\xi\to\infty}\frac{\ln(\xi)}{\xi}\leq{0 }$

This implies by the squeeze theorem that

$\lim_{\xi\to\infty}\frac{\ln(\xi)}{\xi}=0$

And now finally seeing either that

If $\lim_{x\to\infty}|f(x)|=0\text{ that }\lim_{x\to\infty}f(x)=0$

Or more simply that

$\lim_{\xi\to\infty}\frac{-\ln(\xi)}{\xi}=\lim_{\xi\to\infty}-1\cdot\lim_{\xi\to\infty}\frac{\ln(\xi)}{\xi}=-1\cdot{0}=0$

We can finally conclude that

$\lim_{\xi\to\infty}\frac{-\ln(\xi)}{\xi}=0\Rightarrow\lim_{x\to{0}}x\ln(x)=0$

-----------------------------------------------------------------------

Hope that wasn't too formal for you

**flyingsquirrel** · June 28th 2008, 09:25 AM

Hi

Originally Posted by h2osprey

Thanks.
But how would you evaluate

$\lim_{x \to 0^+} x \ln x$

without L'Hôpital's?

Recall the definition of $\ln x$ : $\ln x=\int_1^x\frac{\mathrm{d}t}{t}$

For $t\geq 1$ one has $t \geq \sqrt{t} \implies \frac{1}{t}\leq \frac{1}{\sqrt{t}}\implies \int_1^x\frac{\mathrm{d}t}{t} \leq \int_1^x\frac{\mathrm{d}t}{\sqrt{t}}$

As $\int_1^x\frac{\mathrm{d}t}{\sqrt{t}}= \left[2\sqrt{t} \right]_1^x=2\sqrt{x}-2$ we get $\ln x=\int_1^x\frac{\mathrm{d}t}{t}\leq 2\sqrt{x}-2$

Now let's evaluate $\lim_{x\to \infty} \frac{\ln x}{x}$ using the squeezing theorem. For $x\geq 1$ ,

$0\leq \frac{\ln x}{x} \leq \frac{2\sqrt{x}-2}{x} \Longleftrightarrow 0\leq \frac{\ln x}{x} \leq 2\left( \frac{1}{\sqrt{x}}-\frac{1}{x}\right)$

As $2\left( \frac{1}{\sqrt{x}}-\frac{1}{x}\right) \underset{x\to\infty}{\to}0$ we have $\frac{\ln x}{x} \underset{x\to\infty}{\to}0$ .

Now let $h=\frac{1}{x}$ :

$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{h\to0^+}\frac{\ln \frac{1}{h}}{\frac{1}{h}}=\lim_{h\to0^+} -h\ln h$

Since $\frac{\ln x}{x}\underset{x\to\infty}{\to}0$ we get what we expected : $\lim_{h\to0^+} h\ln h =0$ .

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