1. ## Limits

1. lim {(1+1/x)^x} as x->0

2. lim{(a^x - 1)/x} as x->(+infinity), a>1

Dear friends,
Please try them out without using L'Hopital's rule.

2. Hello, nirmalya basu!

I can help with #1 if you are allowed to use L'Hopital . . .

$\displaystyle 1)\;\;\lim_{x\to0}\left(1 + \frac{1}{x}\right)^x$

Let $\displaystyle y\;=\;\left(1 + \frac{1}{x}\right)^x$

Take logs: .$\displaystyle \ln y\;=\;\ln\left(1 + \frac{1}{x}\right)^x \;= \;x\cdot\ln\left(1 + \frac{1}{x}\right)$

Let $\displaystyle x = \frac{1}{u}$ . . . Note that, if $\displaystyle x\to0$, then $\displaystyle u \to \infty$

We have: .$\displaystyle \ln y \;= \;\frac{1}{u}\ln(1 + u)\;= \;\frac{\ln(1 + u)}{u}$

If $\displaystyle u\to\infty$, we have the indeterminate form: $\displaystyle \frac{\infty}{\infty}$
. . Hence, we may use L'Hopital.

Then we have: .$\displaystyle \ln y\;=\;\frac{\frac{1}{1 + u}}{1}\;=\;\frac{1}{1 + u}$

Hence: .$\displaystyle \lim_{u\to\infty}(\ln y) \;= \;\lim_{u\to\infty}\left(\frac{1}{1 + u}\right)\;=\;0$

Since $\displaystyle \lim_{u\to\infty}(\ln y)\:=\:0$, then $\displaystyle \lim_{u\to\infty} y \:=\:e^0 \:=\:1$

Therefore: .$\displaystyle \lim_{x\to\0}\left(1 + \frac{1}{x}\right)^x\;=\;1$

3. Originally Posted by Soroban
Hello, nirmalya basu!

I can help with #1 if you are allowed to use L'Hopital . . .
There is still another detail missing. You assumed the limit exists! (You probably realized that, right? You just did not mention it cuz to make your work easier to follow).