How do you take this limit algebraically (Not using the graphing calc)

Question

$$\lim_{x\to0}{\frac{e^x-1}{x}}$$

I determined the limit by graphing this and seeing that the graph approaches 1 as x approaches 0. But, is there a way to algebraically determine this limit?

Answer 1

No limit can really be computed by graphing; it's not difficult to present examples where $f(x)$ seems to be close to $1$ when $10^{-n-1}<x<10^{-n}$ (fix $n$ at will) but the limit is $0$. When I was a student, the questions about limits in Calculus exams for the students in Engineering were always of this kind, so they couldn't test with their pocket calculators.

Computing your limit depends on how you defined things. It's equivalent to determine the derivative of the exponential function $x\mapsto e^x$, because $$ \lim_{h\to0}\frac{e^{x+h}-e^x}{h}= \lim_{h\to0}\frac{e^{x}e^{h}-e^x}{h}= \lim_{h\to0}e^x\frac{e^{h}-1}{h}= e^x\lim_{h\to0}\frac{e^{h}-1}{h} $$

So, if you already know this derivative (you might, depending on how you defined the exponential function), you'd know the answer is $1$.

Otherwise, set $e^x-1=1/z$, so that $x=\log(1+1/z)$; when $x$ approaches $0$ from the right, $z$ approaches infinity, so your limit becomes $$ \lim_{x\to0^+}\frac{e^x-1}{x}=\lim_{z\to\infty}\frac{1/z}{\log(1+1/z)} $$ Let's compute the limit of the reciprocal: $$ \lim_{z\to\infty}\frac{\log(1+1/z)}{1/z}= \log\lim_{z\to\infty}\left(1+\frac{1}{z}\right)^z=\log e=1 $$ Similarly you can compute the limit for $x\to0^-$. In conclusion, it depends on the tools you have available and no definitive answer can be given if you don't mention what these tools are.

Answer 2

Hint:

$$ f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}. $$