Why Does My Calculation of the Limit Differ from My Professor's?

  • Thread starter mr_garlic
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I appreciate the reassurance.In summary, the limit as x approaches infinity of the given expression is 0 due to the fact that the bottom term tends towards infinity and the top term remains constant. This is a correct solution and any alternative answers are likely due to starting from a different problem.
  • #1
mr_garlic
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Homework Statement


[tex]\lim_{x\rightarrow\infty} \sqrt{e^{2x}+9}-e^x[/tex]


Homework Equations


[tex]\lim_{x\rightarrow\infty} \sqrt{x} = \infty[/tex]
[tex]\lim_{x\rightarrow\infty} e^x = \infty [/tex]


The Attempt at a Solution



[tex]\lim_{x\rightarrow\infty} \sqrt{e^{2x}+9}-e^x = \lim_{x\rightarrow\infty} \frac{e^{2x}+9-e^{2x}}{\sqrt{e^{2x}+9}+e^x}[/tex]
[tex]=\lim_{x\rightarrow\infty} \frac{9}{\sqrt{e^{2x}+9}+e^x}[/tex]

The limit as x>infty of sqrt(x) is infty
the limit as x>infty of e^2x is infty
the entire bottom term tends towards infinity as x tends towards infinity
therefore
[tex]\lim_{x\rightarrow\infty} \frac{9}{\sqrt{e^{2x}+9}+e^x}=0[/tex]

My professor had a different answer that we didn't have time to go over in class, and my answer just doesn't feel right. Where did I make my mistake?
 
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  • #2
Looks fine to me. I don't see how your professor could come up with a different value unless the two of you started from different problems.
 
  • #3
Thanks, I've been running through this again and again for the past couple days.
 

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