Question

f(x)=2x*e^2x. find lim of f(x) as x--> - infinity and +infinity

Answer 1

\[\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow \infty}2xe ^{2x}\]and\[\lim_{x \rightarrow -\infty}f(x)=\lim_{x \rightarrow -\infty}2xe ^{2x}\]
The first limit is clearly \[+\infty\]since both 2x and e^(2x) approach infinity as x approaches infinity.

The second limit is more difficult since one factor approaches negative infinity and the other approaches 0. To solve this, use L-Hospital's Rule and put e^(2x) in the denominator by making the exponent negative:\[\lim_{x \rightarrow -\infty} 2x/(e ^{-2x})\]Now we have an infinity/infinity situation to which we can apply L'Hospital's Rule. We get\[\lim_{x \rightarrow -\infty}2/(-2e^{-2x}) = \lim_{x \rightarrow -\infty}-e^{2x} = -\infty\]

http://mathworld.wolfram.com/LHospitalsRule.html

Answer 2

you made a mistake

Answer 3

\[\lim_{x \rightarrow -\infty}{2x}{e ^{2x}}\]
\[\lim_{x \rightarrow -\infty}\frac {2x}{e ^{-2x}}\]
\[\lim_{x \rightarrow -\infty}\frac {2}{-2e ^{-2x}}\]=0

Answer 4

Thanks, cinar! I stand corrected. ;-)