1. ## limit..again

allright sorry for asking a million questions..but this one is weird also.
the limit as x goes to 0 ((e^x) - 1) times cosx
the answer in the book is 1 but I always get an undefined answer when I put that limit into a fraction. (either the cosx on the bottom of the fraction..((e^x) - 1)^-1/cosx..or secx/e^x - 1 both give me invalid answers.) Thx for the help.

edit..ok I looked online for the mistakes in book. It should be ((e^x-1) times cotx

2. $\displaystyle \displaystyle \lim_{x\to 0}(e^x-1)\cot x=\lim_{x\to 0}\frac{e^x-1}{\tan x}=\lim_{x\to 0}\frac{e^x-1}{x}\cdot\frac{x}{\tan x}=1\cdot 1=1$

3. Originally Posted by red_dog
$\displaystyle \displaystyle \lim_{x\to 0}(e^x-1)\cot x=\lim_{x\to 0}\frac{e^x-1}{\tan x}=\lim_{x\to 0}\frac{e^x-1}{x}\cdot\frac{x}{\tan x}=1\cdot 1=1$

I saw you broke that up into two fractions. How does that work and why did you do that?

4. If $\displaystyle \displaystyle \lim_{x\to x_0}f(x)=l_1$ and $\displaystyle \displaystyle \lim_{x\to x_0}g(x)=l_2$ then $\displaystyle \displaystyle \lim_{x\to x_0}f(x)g(x)=\displaystyle \lim_{x\to x_0}f(x)\cdot \displaystyle \lim_{x\to x_0}g(x)=l_1\cdot l_2$. (Exception $\displaystyle 0\cdot\infty)$.

$\displaystyle \displaystyle \lim_{x\to 0}\frac{e^x-1}{x}=1$ by l'Hospital
and $\displaystyle \displaystyle \lim_{x\to 0}\frac{x}{\tan x}=1$ is a remarcable limit.