# Thread: Find the limit

1. ## Find the limit

Question : Find the Limite $\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}$

My Work :::::::::::::::::::::::::::::::::::::::::::

$((e^x - 1) + (1+x))^{\frac{1}{x}}$.....................Is this step correct

2. As, you usually say to me, I have no idea what you are doing.
The idea here is to take the logarithm and then use L'Hopital's rule.

3. Originally Posted by zorro
Question : Find the Limite $\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}$

My Work :::::::::::::::::::::::::::::::::::::::::::

$((e^x - 1) + (1+x))^{\frac{1}{x}}$.....................Is this step correct
Two main methods:

Method one: $\lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x}}= \lim_{x\to\infty}e\left(1+\frac{x}{e^x}\right)^{\f rac{1}{x}}$ and it is quite easy to validate that $\lim_{x\to\infty}\left(1+\frac{x}{e^x}\right)^{\fr ac{1}{x}}=\exp\left(\lim_{x\to\infty}\frac{\ln\lef t(1+\tfrac{x}{e^x}\right)}{x}\right)$ and since $\ln\left(1+\tfrac{x}{e^x}\right)\underset{x\to\inf ty}{\sim}\tfrac{x}{e^x}$ it follows that our limit is equal to $\exp\left(\lim_{x\to\infty}\frac{\tfrac{x}{e^x}}{x }\right)=e^0=1$ so then the actual limit is clearly $e$.

Alternatively,

$e\leqslant\left(e^x+x\right)^{\frac{1}{x}}\leqslan t 2^{\frac{1}{x}}e$. And since $\lim_{x\to\infty}2^{\frac{1}{x}}=1$ the limit follows by the squeeze theorem.

4. I just let

$y=\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}$

then take the log of y, pass through the limit and apply L'Hopital rule's.

5. Originally Posted by matheagle
I just let

$y=\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}$

then take the log of y, pass through the limit and apply L'Hopital rule's.
That is of course equally valid!

6. $y = (e^x + x)^\frac{1}{x}$

taking log we get

$= \frac{ln (e^x+x)}{x}$

Using l'Hospital rule

$f'(x) = \frac{d}{dx} \left[ ln(e^x + x) \right]$ = $\frac{e^x + 1}{e^x+x}$

$g'(x) = 1$

$\lim_{x \to 0} \frac{f'(x)}{g'(x)}$ = $\lim_{x \to 0} \ \frac{e^x+1}{e^x+x}$ .............what should i do know???

7. Originally Posted by Drexel28

$
\lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x}}= \lim_{x\to\infty}e\left(1+\frac{x}{e^x}\right)^{\f rac{1}{x}}
$

A nice method !! How could you think of this ...

You did the limit for $x \to \infty$ ,
now if $x \to 0$ , let me finish it

$\lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

$= e \lim_{x\to 0} \left( 1 + \frac{ e^{-x}}{ \frac{1}{x}} \right)^{\frac{1}{x}} = e \lim_{x\to 0} e^{e^{-x}} = e^2$

8. Here is another method to reach the limit but $x \to \infty$

starting from

$e \lim_{x\to \infty} \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

which is the result from Drexel28 .

since $\ln(u) = \lim_{a\to 0} \frac{ u^a - 1}{a}$

$\lim_{x\to \infty} \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

$= \lim_{x\to\infty} \frac{1}{x} \cdot \frac{\left [ \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}} - 1 \right ]}{ \frac{1}{x}} + 1$

sub $\frac{1}{x} = a$

$= \lim_{a\to 0} a \ln( 1 + \frac{{e^{- \frac{1}{a}}}}{a} ) + 1$

we can see that it equals to $1$

( $0 * \ln(1) + 1$ )

9. Originally Posted by simplependulum
A nice method !! How could you think of this ...

You did the limit for $x \to \infty$ ,
now if $x \to 0$ , let me finish it

$\lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

$= e \lim_{x\to 0} \left( 1 + \frac{ e^{-x}}{ \frac{1}{x}} \right)^{\frac{1}{x}} = e \lim_{x\to 0} e^{e^{-x}} = e^2$
Another really easy thing one can prove is that let $L=\left(P(x)+\alpha^x+\beta^x+\cdots\zeta^x\right) ^{\frac{1}{x}}$ be such that $P(x)$ is an arbitrary polynomial and $0<\alpha<\beta<\cdots<\zeta$ then $\ln\text{ }L=\lim_{x\to\infty}\frac{\ln\left(P(x)+\alpha^x+\ beta^x+\cdots+\zeta^x\right)}{x}$ $=\lim_{x\to\infty}\left\{\ln\left(\zeta\right)+\fr ac{\ln\left(\tfrac{P(x)}{\zeta^x}+\left(\tfrac{\al pha}{\zeta}\right)^x+\left(\tfrac{\beta}{\zeta}\ri ght)+
\cdots+1\right)}{x}\right\}$
and since $\frac{P(x)}{\zeta^x}+\left(\frac{\alpha}{\zeta}\ri ght)^x+\left(\frac{\beta}{\zeta}\right)^x+\cdots\t o0$ we can see that $\ln\left(\frac{P(x)}{\zeta^x}+\left(\tfrac{\alpha} {\zeta}\right)^x+\left(\tfrac{\beta}{\zeta}\right) ^x+\cdots+1\right)\underset{x\to\infty}{\sim}\frac {P(x)}{\zeta^x}+\left(\frac{\alpha}{\zeta}\right)^ x+\left(\frac{\beta}{\zeta}\right)^x+\cdots$ from where it clearly follows that $\ln\text{ }L=\ln\left(\zeta\right)\implies L=\zeta$. A similar proof can be done wit h the squeeze theorem.

This can be GREATLY generalized.

10. ## shouldnt it be this

Originally Posted by simplependulum
A nice method !! How could you think of this ...

You did the limit for $x \to \infty$ ,
now if $x \to 0$ , let me finish it

$\lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

I am no genuis but should it be this

$\lim_{x\to 0} e^x \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

u have taken $e$ out side instead of $e^x$

11. Originally Posted by zorro
I am no genuis but should it be this

$\lim_{x\to 0} e^x \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}$

u have taken $e$ out side instead of $e^x$
Yes, you are right. It was probably a typo.

12. Originally Posted by zorro
$y = (e^x + x)^\frac{1}{x}$

taking log we get

$= \frac{ln (e^x+x)}{x}$

Using l'Hospital rule

$f'(x) = \frac{d}{dx} \left[ ln(e^x + x) \right]$ = $\frac{e^x + 1}{e^x+x}$

$g'(x) = 1$

$\lim_{x \to 0} \frac{f'(x)}{g'(x)}$ = $\lim_{x \to 0} \ \frac{e^x+1}{e^x+x}$ .............what should i do know???
Now just set x= 0!