# Calculate fundamental limits using l'Hospital rule

So I have this essay where a question is "Calculate the three fundamental limits using l'Hospital's rule"

I find easy to calculate $\lim_{x \rightarrow 0}\frac{\sin(x)}{x}$ and $\lim_{x \rightarrow 0}\frac{e^x - 1}{x}$, however the one I can't understand is the limit $\lim_{x \rightarrow +\infty}\left(1 + \frac{1}{x}\right)^x$... How exactly am I supposed to use l'Hospital's rule here?

I tried writing $\left(1 + \frac{1}{x}\right)^x$ as $\frac{(x+1)^x}{x^x}$ and utilize the fact that $\frac{d(x^x)}{dx} = x^x(ln(x) + 1)$ but instead of simplifying, using l'Hospital'a rule that way actually makes it worse...

Can anyone point me to the right direction?

• I presume that sen(x) means \sin x – DanielWainfleet Aug 9 '18 at 1:10
• A strange topic for an essay, since using l'Hospital for those limits is circular reasoning (with the way that the derivatives of $\sin x$ and $e^x$ are usually derived in basic calculus courses)... – Hans Lundmark Aug 9 '18 at 10:57

## 2 Answers

HINT

By the well known exponential manipulation $A^B=e^{B\log A}$, we have

$$\left(1 + \frac{1}{x}\right)^x=\large{e^{x\log \left(1 + \frac{1}{x}\right)}}=\large{e^{\frac{\log \left(1 + \frac{1}{x}\right)}{\frac1x}}}$$

and $\frac{\log \left(1 + \frac{1}{x}\right)}{\frac1x}$ is an indeterminate form $\frac{0}{0}$.

Hint: $(1+1/x)^x=e^{x \ln(1+1/x)} = e^{\ln(1+t)/t}$ where $t=1/x$.