What is the vertical asymptote for g(x) = e^(-1/x^2) using limits?

Question

ganeshie8 · Answer

solve for $a$ below :-

$\large \lim \limits_{x \to a} g(x) = \pm\infty$

anonymous · Answer

I'm not sure how to solve for a...But I do know that it is undefined at x=0, so I tried doing Lim x--> 0 g(x) = 0?

ganeshie8 · Answer

yeah, so no VA i think; eventhough it is undefined. seeing the graph makes everything clear

ganeshie8 · Answer

there is no $a$ for which the graph goes +- infinity, so no VA

anonymous · Answer

Ok! When I graphed in on my calculator though, it looked like it never intersects x=0?

ganeshie8 · Answer

$e^{-1/x^2} = 1/e^{1/x^2}$
as x->0, 1/x^2 -> infinity
bottom -> infinity

g(x) -> 0

but since 0 is not in domain, graph wont be there at that point

ganeshie8 · Answer

function being undefined is a necessary condition for VA. Not a sufficient condition. in this particular example, the VA doesnt exist

anonymous · Answer

Ok! I understand it! That's a great explanation.

ganeshie8 · Answer

glad to hear :)