# Limits

• Aug 3rd 2006, 05:10 AM
nirmalya basu
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.
• Aug 3rd 2006, 06:04 AM
Soroban
Hello, nirmalya basu!

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

Quote:

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

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

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

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

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

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

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

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

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

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

• Aug 3rd 2006, 08:45 AM
ThePerfectHacker
Quote:

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).