Thanks so much that's a very nice way to approach it!\(\displaystyle \dfrac {1}{e^{-\infty}}\) tends to infinity, NOT 0.
I always like to change these type of problems from x goes to -infinty to x goes to +infinity AND change all x's to -x[/tex]
Also, there is no need to trust 'solution' as it may be wrong.
All you need is to replace x with say -10,000 and see if that result is close to -2/3, -3 or something else.
Thank you. I guess I need to be careful about weather it's negative or positive cuz it's a totally different questions eachYou're thinking in terms of [imath]x[/imath] trending toward *positive* infinity. But the limit is for *negative* infinity.
When you're on the negative side of the [imath]x[/imath]-axis, [imath]e^{2x}[/imath] is going to get very small, because the [imath]2[/imath] is positive and the [imath]x[/imath] is getting strongly negative, so the base is being raised to a negative power. So it is actually the [imath]e^x[/imath] terms that are disappearing, leaving you with [imath]\frac{-3}{1}[/imath].
I guess I need to be careful about weather it's negative or positive cuz it's a totally different questions each