# How do you find the limit of (1-e^(-2x))/(sec(x)) as x approaches 0?

Mar 8, 2016

$0$

#### Explanation:

Step 1:
$\frac{1 - {e}^{- 2 x}}{\sec} \left(x\right) \rightarrow \left(1 - {e}^{- 2 x}\right) \cos \left(x\right)$

because $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

Step 2:
the cosine function is known to have a range of $- 1 \le \cos \left(x\right) \le 1$

So we can write:
$- 1 \left(1 - {e}^{- 2 x}\right) \le \left(1 - {e}^{- 2 x}\right) \cos \left(x\right) \le 1 \left(1 - {e}^{- 2 x}\right)$

Step 3:
${\lim}_{x \rightarrow 0} - \left(1 - {e}^{- 2 x}\right) \rightarrow {\lim}_{x \rightarrow 0} - 1 + \frac{1}{e} ^ \left(2 x\right) = - 1 + \frac{1}{e} ^ \left(0\right) = - 1 + \frac{1}{1} = 0$

Step 4:
${\lim}_{x \rightarrow 0} \left(1 - {e}^{- 2 x}\right) \rightarrow {\lim}_{x \rightarrow 0} 1 - \frac{1}{e} ^ \left(2 x\right) = 1 - \frac{1}{e} ^ \left(0\right) = 1 - \frac{1}{1} = 0$

Hence, ${\lim}_{x \rightarrow 0} \left(1 - {e}^{- 2 x}\right) \cos \left(x\right) = \textcolor{b l u e}{0}$

Recall:
According to the color(black)("Sandwich theorem"), if $a \left(x\right) \le f \left(x\right) \le b \left(x\right)$
And ${\lim}_{x \rightarrow c} a \left(x\right) = K = {\lim}_{x \rightarrow c} b \left(x\right)$
this implies that ${\lim}_{x \rightarrow c} f \left(x\right) = K$