Question

Find the limit if it exists. Use L' Hospital rule.

Answer 1

\[\lim_{x \rightarrow 0}(\frac{ 1 }{ x }-\frac{ 1 }{ e^x-1 })\]

Answer 2

what do you get when you combine the fractions?

Answer 3

\[\lim_{x \rightarrow 0}(\frac{ (e^x-1 )-x}{ x(e^x-1) }\]

Answer 4

I'm getting the same

Answer 5

if you plug in x = 0, you should get 0/0 which is one of the many indeterminate forms

so use L'Hospital's rule to get what?

Answer 6

do I simplify the bottom first then use the L'hospital rule?

Answer 7

simplify how?

Answer 8

\[xe^x-x\]

Answer 9

sure you can distribute

Answer 10

\[\lim_{x \rightarrow 0}\frac{ (e^x-1)-1 }{ xe^x \times e^x-1 }\]

Answer 11

then use L'Hospital rule again?

Answer 12

after using L'Hospital rule the first time, you should have

\[\Large \lim_{x \to 0}\frac{e^x-1}{e^x+xe^x-1}\]

Answer 13

plugging in x = 0 into that will lead to 0/0 again

so you have to use L'Hospital's rule again

Answer 14

but isn't \[(e^x-1)-x = e^x-1\] since (-x) = -1?

Answer 15

\[(e^x-1)-1\]**

Answer 16

taking the derivative of -1 is 0

Answer 17

oh you did the L'hospital rule the 2nd time then.

Answer 18

Original
\[\Large \lim_{x \to 0}\frac{ (e^x-1 )-x}{ x(e^x-1) }\]

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after applying L'Hospital's rule the first time, you'll get
\[\Large \lim_{x \to 0}\frac{e^x-1}{e^x+xe^x-1}\]

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after applying L'Hospital's rule the second time, you'll get
\[\Large \lim_{x \to 0}\frac{e^x}{2e^x+xe^x}\]

Answer 19

then since top and bottom isn't 0/0, we can plug in 0 to get 1/2?

Answer 20

1/2 is your final answer, yep

Answer 21

okay, got it. Thanks! :D

Answer 22

np