Saturday 18 June 2016

calculus - 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?


Answer



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 $\frac{0}{0}$.



No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...