Question

steps on how to evaluate this limit

lim as x approaches 0

(e^x) -1/ sin(13x)

Answer 1

Step #1 is to write clearly. Do you REALLY mean what you have written? \((e^{x}) - \dfrac{1}{\sin(13x)}\)

Or did you mean \(\dfrac{e^{x} - 1}{\sin(13x)}\)?

Answer 2

oh, i'm sorry i should i have written it more clearly. the bottom one! sorry about that..

Answer 3

have you tried l'hospital's rule?

Answer 4

that's the part i don't understand how to do.. :(

Answer 5

awh, ok
do you know how to take the derivative?

Answer 6

First, make sure we have an Indeterminate Form. Don't just use L'Hospital's rule willy-nilly.

Answer 7

I'm tempted just to rewrite it. \(\dfrac{e^{x}-1}{\sin(13x)} =\dfrac{13x}{\sin(13x)}\cdot\dfrac{e^{x} - 1}{13x}\)

The expression on the right is quite a bit easier to fathom.

Answer 8

would i take the derivative of the expression on the right?

Answer 9

I'd just take the derivative of what they gave you;
e^x - 1 and
sin(13x)

Answer 10

d (e^x - 1)dx = ?

Answer 11

would it be e^x ?

Answer 12

for sin(13x) would the derivative be cos(13x) ?

Answer 13

e^x is correct
for sin(13x) you'll need to use 'u-substitution' it would look like this:
d(sin(13x))dx,
let u = 13x,
then
du = 13,
and
d(sin(u)) = cos(u)du
cos(u)du = cos(13x)*13 or 13cos(13x)

Answer 14

now take the limit of this:
\[\lim_{x \rightarrow 0}\frac{ e^x }{ 13\cos(13x) }\]

Answer 15

would it be 1/13 ?

Answer 16

Under what circumstances would it be something else?

Answer 17

haha i just wanted to make sure... thank you everyone!