What is the limit of (e^x-1)/x as x approaches 0?

[kidia]

friends help on this,

Use the ineequalities 1+x<=e^x<=1/1-x for |x|<1 to show that
lim(e^x-1)/x =
x-0+
lim(e^x-1)/x =1
x-0-
and hence deduce that
lim(e^x-1)/x=1
x-0
b) Determine
lim (x^3-1)/(x-1) if it exists.
x-1

 

[Galileo]

Rewrite your inequalities.
For example:

$$1+x \leq e^x$$

is the same as:

$$1\leq \frac{e^x-1}{x}$$

Do the same for the other side of the inequality and use the squeeze theorem to evaluate the limit.

 

[arildno]

As for b), perform polynomial division first.
Or L'Hopital's rule if you're allowed to do so.