# Math Help - Evaluation of limit (2)

1. ## Evaluation of limit (2)

pliz help me out with this,

$lim_{x\rightarrow 0} \frac{(1+x)^{1/x}-e}{x}$

2. ## Re: Evaluation of limit (2)

what happens if you make the substitution n = 1/x?

3. ## Re: Evaluation of limit (2)

ok..after replacing n=1/x....i gt
$lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}$...
what to do next...

4. ## Re: Evaluation of limit (2)

Originally Posted by rishilaish
ok..after replacing n=1/x....i gt
$lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}$...
what to do next...
$\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^n = e$

This is a common limit well worth knowing.

Edit: you need to change your denominator too: $n = \dfrac{1}{x} \Leftrightarrow x = \dfrac{1}{n}$

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

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

5. ## Re: Evaluation of limit (2)

Originally Posted by rishilaish
ok..after replacing n=1/x....i gt
$lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}$...
what to do next...
It would actually be $\displaystyle \lim_{x \to \infty}\frac{\left(1 + \frac{1}{n}\right)^n - e}{\frac{1}{n}}$

Personally I don't see making the substitution $\displaystyle n = \frac{1}{x}$ as necessary, since the function already went to $\displaystyle \frac{0}{0}$ and the denominator of the original function is easier to differentiate so that you can use L'Hospital's Rule...

6. ## Re: Evaluation of limit (2)

Originally Posted by e^(i*pi)
$\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^n = e$

This is a common limit well worth knowing.

Edit: you need to change your denominator too: $n = \dfrac{1}{x} \Leftrightarrow x = \dfrac{1}{n}$

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

$= \lim_{n \to \infty} \left(n \left(1+\frac{1}{n}\right)^n - e\right)$
i think

$\lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e$

7. ## Re: Evaluation of limit (2)

Originally Posted by rishilaish
i think

$\lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e$

No, $n$ has to approach to $\infty$. But this limit is not easy, because if you follow the suggestion of Prove It by applying l'Hopital's rule then the derivative of $(1+x)^{\frac{1}{x}}$ is not very easy and I don't know if it simplifies the problem.

8. ## Re: Evaluation of limit (2)

Originally Posted by rishilaish
i think

$\lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e$

That's not the case, it is definitely the limit to infinity: lim n to infinity &#40;1&#43;1&#47;n&#41;&#94;n - Wolfram|Alpha

If you want choose a suitable large value of $n$ and compare it to e.

9. ## Re: Evaluation of limit (2)

somebody plz solve it...the answer is -e/2

10. ## Re: Evaluation of limit (2)

By first inspection you notice you get the indeterminate form $\frac{0}{0}$ therefore you can apply l'Hopital's rule, that means:
$\lim_{x \to 0} \frac{\frac{d}{dx}(1+x)^{\frac{1}{x}}-\frac{d(e)}{dx}}{\frac{d(x)}{dx}}$
$=\lim_{x\to 0} \frac{d}{dx}(1+x)^{\frac{1}{x}}$

Calculating $\frac{d}{dx}(1+x)^{\frac{1}{x}}$ gives:
$\frac{d}{dx}e^{\frac{\ln(1+x)}{x}}$
$=e^{\frac{\ln(1+x)}{x}}\cdot \frac{d}{dx}\left[\frac{\ln(1+x)}{x}\right]$

Calculating $\frac{d}{dx}\left[\frac{\ln(1+x)}{x}\right]$ gives:
$\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$

So now you have this limit:
$\lim_{x\to 0} e^{\frac{\ln(1+x)}{x}}\cdot \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$

which is equal to (because the limit of a product is the product of the limits of the separated factors):
$\lim_{x\to 0} e^{\frac{\ln(1+x)}{x}} \cdot \lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$

And
$\lim_{x\to 0} e^{\frac{\ln(1+x)}{x}}=e$ (recheck that!)

Therefore the only thing we have to do is to calculate:
$\lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$
You notice you get again the indeterminate form $\frac{0}{0}$ therefore we can apply l'Hopital's rule again:
$\lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}=\lim_{x\to 0} \frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}$
$=\lim_{x \to 0} \frac{-x}{(1+x)^2\cdot 2x}=\lim_{x\to 0}\frac{-1}{2(1+x)^2}=\frac{-1}{2}$

Therefore the limit is $\frac{-e}{2}$

11. ## Re: Evaluation of limit (2)

thanks siron...u help me out every detail...examz getting close so iv to prepare hard...