Question

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

Answer 1

please let me know if you used l'hospital's rule or not

Answer 2

Why don't you decide if l'hospital is applicable and then apply the rule?

Answer 3

in the case of the first, I'm not sure how the answer is 0

Answer 4

Did you use l'hospital? What results came for your attempt at a solution?

Answer 5

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?

Answer 6

\[\lim_{x \rightarrow -\infty}\frac{ e^3x }{ x^2+1 }\]
divide by x^2 \[\lim_{x \rightarrow -\infty}\frac{ \frac{ e^3 }{ x } }{ 1+\frac{ 1 }{ x^2 } }\]
x=-infinity
anything/x =0\[\lim_{x \rightarrow -\infty}\frac{ 0 }{ 1+0 }=0\]

Answer 7

note these 2 points\[\frac{ anything }{ \infty }=0\]\[\frac{ anything }{-\infty }=0\]

Answer 8

does it matter at all that its

\[\lim_{x \rightarrow -\infty} \frac{ e^{3x} }{ x^2+1 }\]

Answer 9

ok forget what all i did up there^

Answer 10

\[\lim_{x \rightarrow -\infty}\frac{ e^{3x} }{ (x^2+1) }\]\[\lim_{x \rightarrow -\infty}\frac{ e^{3\times -\infty} }{ (-\infty)^2+1}\]\[\lim_{x \rightarrow -\infty}\frac{ e^{-\infty} }{ \infty^2 +1 }=\lim_{x \rightarrow -\infty}\frac{ 0 }{ \infty }=0\]

Answer 11

why is \[e^{-\infty} \] = 0?

Answer 12

wait nevermind it becomes 1/e to the negative infinity. thank you!

Answer 13

yes :) np