anonymous
  • anonymous
Find the following limits: the limit of (e^3x)/(x^2+1) as the limit approaches negative infinity the limit of x ln(x) as the limit approaches 0 from the right
Calculus1
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
please let me know if you used l'hospital's rule or not
tkhunny
  • tkhunny
Why don't you decide if l'hospital is applicable and then apply the rule?
anonymous
  • anonymous
in the case of the first, I'm not sure how the answer is 0

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tkhunny
  • tkhunny
Did you use l'hospital? What results came for your attempt at a solution?
anonymous
  • anonymous
I know when it reaches positive infinity, the answer DNE because since e is to the power of 3x, it would be approaching a larger, infinite number and deriving x^2+1 would eventually give me 2. But I'm not sure why the answer is 0 in this case? How does it change because it is approaching negative infinity?
anonymous
  • anonymous
\[\lim_{x \rightarrow -\infty}\frac{ e^3x }{ x^2+1 }\] divide by x^2 \[\lim_{x \rightarrow -\infty}\frac{ \frac{ e^3 }{ x } }{ 1+\frac{ 1 }{ x^2 } }\] x=-infinity anything/x =0\[\lim_{x \rightarrow -\infty}\frac{ 0 }{ 1+0 }=0\]
anonymous
  • anonymous
note these 2 points\[\frac{ anything }{ \infty }=0\]\[\frac{ anything }{-\infty }=0\]
anonymous
  • anonymous
does it matter at all that its \[\lim_{x \rightarrow -\infty} \frac{ e^{3x} }{ x^2+1 }\]
anonymous
  • anonymous
ok forget what all i did up there^
anonymous
  • anonymous
\[\lim_{x \rightarrow -\infty}\frac{ e^{3x} }{ (x^2+1) }\]\[\lim_{x \rightarrow -\infty}\frac{ e^{3\times -\infty} }{ (-\infty)^2+1}\]\[\lim_{x \rightarrow -\infty}\frac{ e^{-\infty} }{ \infty^2 +1 }=\lim_{x \rightarrow -\infty}\frac{ 0 }{ \infty }=0\]
anonymous
  • anonymous
why is \[e^{-\infty} \] = 0?
anonymous
  • anonymous
wait nevermind it becomes 1/e to the negative infinity. thank you!
anonymous
  • anonymous
yes :) np

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