1. Question on Limits

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.

2. 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 $\displaystyle \frac{\infty}{\infty}$ or $\displaystyle \frac 00$...

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

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

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

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

3. 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.
$\displaystyle \color{red}0^0\ne{1}$

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

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

So we have

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

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

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

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

4. Originally Posted by Moo
Hello,

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

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

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

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

---> $\displaystyle \lim_{x \to 0^+} x^{\tan x}=\lim_{x \to 0^+} e^{\tan x \ln x}=\lim_{x \to 0^+} e^{x \ln x}=e^0=1$
Thanks.
But how would you evaluate

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

without L'Hôpital's?

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

6. Originally Posted by h2osprey
Thanks.
But how would you evaluate lim x->0+ for xlnx without L'Hopital's?
$\displaystyle \lim_{x\to{0}}x\ln(x)$

There are many ways, but the easies would probably be

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

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

So we would have

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

Now we can say one of two things

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

Or

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

And

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

This implies that

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

Now remaining inequality

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

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

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

This implies by the squeeze theorem that

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

And now finally seeing either that

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

Or more simply that

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

We can finally conclude that

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

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Hope that wasn't too formal for you

7. Hi
Originally Posted by h2osprey
Thanks.
But how would you evaluate

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

without L'Hôpital's?
Recall the definition of $\displaystyle \ln x$ : $\displaystyle \ln x=\int_1^x\frac{\mathrm{d}t}{t}$

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

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

Now let's evaluate $\displaystyle \lim_{x\to \infty} \frac{\ln x}{x}$ using the squeezing theorem. For $\displaystyle x\geq 1$,
$\displaystyle 0\leq \frac{\ln x}{x} \leq \frac{2\sqrt{x}-2}{x} \Longleftrightarrow 0\leq \frac{\ln x}{x} \leq 2\left( \frac{1}{\sqrt{x}}-\frac{1}{x}\right)$
As $\displaystyle 2\left( \frac{1}{\sqrt{x}}-\frac{1}{x}\right) \underset{x\to\infty}{\to}0$ we have $\displaystyle \frac{\ln x}{x} \underset{x\to\infty}{\to}0$.

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

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