# Step-by-step Solution

## Find the limit of $\frac{1-e^x}{\tan\left(x\right)}$ as $x$ approaches $0$

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
÷
2

e
π
ln
log
log
lim
d/dx
Dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

### Videos

$-1$

## Step-by-step explanation

Problem to solve:

$\lim_{x\to0}\left(\frac{\left(1-e^x\right)}{\tan\left(x\right)}\right)$
1

As the limit results in indeterminate form, we can apply L'Hôpital's rule

$\lim_{x\to0}\left(\frac{\frac{d}{dx}\left(1-e^x\right)}{\frac{d}{dx}\left(\tan\left(x\right)\right)}\right)$
2

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

$\lim_{x\to0}\left(\frac{\frac{d}{dx}\left(1-e^x\right)}{\sec\left(x\right)^2\frac{d}{dx}\left(x\right)}\right)$

$-1$
$\lim_{x\to0}\left(\frac{\left(1-e^x\right)}{\tan\left(x\right)}\right)$

Limits

~ 1.43 seconds