Question

limit as x -> 0 of (x cos x + e^-x)/x^2

Answer 1

\[=\frac{ \lim_{x \rightarrow 0}(x \cos x + e^{-x}) }{ \lim_{x \rightarrow 0}(x^2)}\]
\[=\frac{ \lim_{x \rightarrow 0}(x \cos x) + \lim_{x \rightarrow 0}(e^{-x}) }{ 0 }\]
\[=\frac{ \lim_{x \rightarrow 0}(x) \lim_{x \rightarrow 0}(\cos x) + 0 }{ 0 }\]
\[=\frac{ 0 \lim_{x \rightarrow 0}(\cos x) + 0 }{ 0 }\]
\[=\frac{ 0 }{ 0 }\]
Indeterminate Form

Answer 2

Somehow the answer in my book says +infinity. Is it because since it is a indeterminate form do you have to apply L' Hopitals rule?

Answer 3

Yea, that's what you would do :)

Answer 4

So do you have to keep on doing the derivative until the denominator is a 2?

Answer 5

Yes. The numerator at that point should be +infinity

Answer 6

@DDCamp, \(\displaystyle\lim_{x\to0}e^{-x}=1\), not 0.

Answer 7

Oh, I guess it is. For some reason I did lim(x→infinity) for that part.

This break has been too long...

Answer 8

So how do I solve this if its now 1/0?

Answer 9

1/0 would just be infinity.

Answer 10

Oh... sorry its been a long day. I feel like an idiot.

Answer 11

You still have to be careful about that conclusion. For example, for \(\dfrac{1}{x}\), as \(x\to0\), the two-sided limit doesn't exist.