Question

Evaluate this limit

Answer 1

do it bud

Answer 2

post it!

Answer 3

\[\lim_{x \rightarrow \pi} \frac{ e^{\sin(x)}-1 }{ x-\pi }\]

clearly just plugging in x = pi wont work. because you would get zero in the denominator.

This is what I waS thinking.

\[\frac{ \frac{ d }{ dx }e^{\sin(x)}-1 }{ \frac{ d }{ dx}x-\pi } = e^{\sin(x)}*\cos(x) \]

Answer 4

i'm getting zero for this

Answer 5

\[e^{\sin(\pi}*Cos(\pi) = -1*0 = 0\]

Answer 6

Use Lhopital's rule

Answer 7

cos(pi) is not 0

Answer 8

e^(sin(pi)) is also not 0

Answer 9

so not sure where you get 0 from

Answer 10

maybe you know cos(pi) is -1
but e^(sin(pi)) is e^0 which is 1

Answer 11

oh. totally missed that

Answer 12

\[\sin(\pi) = 0, e^(0) = 1 \]

\[-1+1 =0 \]

Answer 13

$$ \lim_{x \rightarrow \pi} \frac{ e^{\sin(x)}-1 }{ x-\pi }
\implies \lim_{x \rightarrow \pi} \frac{ e^{\sin(x)} \cos x }{ 1 }

$$

Answer 14

should be multiplication not addition

Answer 15

i've been doing problems all night lol... i think i need to take a break.

Answer 16

\[e^{\sin(\pi)} \cdot \cos(\pi)=e^{0} \cdot (-1)=1 (-1)=-1\]