anonymous
  • anonymous
limit as x -> 0 of (x cos x + e^-x)/x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DDCamp
  • DDCamp
\[=\frac{ \lim_{x \rightarrow 0}(x \cos x + e^{-x}) }{ \lim_{x \rightarrow 0}(x^2)}\] \[=\frac{ \lim_{x \rightarrow 0}(x \cos x) + \lim_{x \rightarrow 0}(e^{-x}) }{ 0 }\] \[=\frac{ \lim_{x \rightarrow 0}(x) \lim_{x \rightarrow 0}(\cos x) + 0 }{ 0 }\] \[=\frac{ 0 \lim_{x \rightarrow 0}(\cos x) + 0 }{ 0 }\] \[=\frac{ 0 }{ 0 }\] Indeterminate Form
anonymous
  • anonymous
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?
DDCamp
  • DDCamp
Yea, that's what you would do :)

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anonymous
  • anonymous
So do you have to keep on doing the derivative until the denominator is a 2?
DDCamp
  • DDCamp
Yes. The numerator at that point should be +infinity
anonymous
  • anonymous
@DDCamp, \(\displaystyle\lim_{x\to0}e^{-x}=1\), not 0.
DDCamp
  • DDCamp
Oh, I guess it is. For some reason I did lim(x→infinity) for that part. This break has been too long...
anonymous
  • anonymous
So how do I solve this if its now 1/0?
DDCamp
  • DDCamp
1/0 would just be infinity.
anonymous
  • anonymous
Oh... sorry its been a long day. I feel like an idiot.
anonymous
  • anonymous
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.

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