Question #ce538

The limit differs from either side, so the limit is undefined.

graph{y = (1/x^2)/(e^(-1/x)) [-8.315, 9.47, -4.95, 3.936]}

DISCLAIMER: LONG ANSWER!

The limit originally is:

#L = lim_(x->0) (1/(x^2))/(e^(-1//x)) = lim_(x->0) 1/(x^2e^(-1//x))#

Since #0cdot0 = 0#, this is of the form #1/oo#, so we could try evaluating this individually for #x^2# and for #e^(-1//x)# to break this problem down.

#L = 1/(lim_(x->0) x^2 cdot lim_(x->0) e^(-1//x))#

The first term is easy:

#color(green)(lim_(x->0) x^2 = 0)#
(trivial)

Now for the second term.

LIMIT FROM THE LEFT SIDE

And now, we evaluate the limit from the left side of #e^(-1//x)#.

#L_(-,e^(-1//x)) = lim_(x->0^(-)) e^(-1//x)#

#ln L_(-,e^(-1//x)) = ln[lim_(x->0^(-)) e^(-1//x)]#

This limit is continuous in the open interval before reaching #0# from either side, so we can say:

#ln L_(-,e^(-1//x)) = lim_(x->0^(-)) ln e^(-1//x)#

#= -lim_(x->0^(-)) 1/x#

#= -(-oo) = +oo#

Thus, #color(green)(L_(-,e^(-1//x)) = "exp"(-lim_(x->0^(-)) 1/x) = e^(+oo) = oo)#.

From this, we have the overall limit from the left side is:

#L_(-) = lim_(x->0^(-)) 1/(x^2e^(-1//x))#

#= 1/(0 cdot L_(-,e^(-1//x))) prop 1/(0*1/0) prop 0/0#

From the negative side, we can thus use L'Hopital's Rule:

#L_(-) = lim_(x->0^(-)) e^(1//x)/(x^2)#

This is going to work better if we take the #ln#:

#ln L_(-) = lim_(x->0^(-)) ln (e^(1//x)/(x^2))#

#= lim_(x->0^(-)) (1/x - ln x^2)#

#= lim_(x->0^(-)) (1 - xln x^2)/x#

#= lim_(x->0^(-)) (0 - cancel(x) cdot 1/cancel(x^2) cdot 2cancel(x) + ln x^2)/1#

By the symmetry of #lnx^2# about the #y# axis, the limit from the left is equal to the limit from the right, which means...

#ln L_(-) = lim_(x->0^(-)) (-2 + ln x^2)/1 = -oo#

Therefore, #bb(L_(-) = e^(-oo) = 0)#. Almost there!

LIMIT FROM THE RIGHT SIDE

Now for the limit from the right side of #e^(-1//x)#.

#L_(+,e^(-1//x)) = lim_(x->0^(+)) e^(-1//x)#

#ln L_(+,e^(-1//x)) = ln[lim_(x->0^(+)) e^(-1//x)]#

As before...

#= lim_(x->0^(+)) ln e^(-1//x)#

#= -lim_(x->0^(+)) 1/x#

#= -(+oo) = -oo#

Thus, #color(green)(L_(+,e^(-1//x)) = "exp"(-lim_(x->0^(+)) 1/x) = e^(-oo) = 0)#.

For the overall limit from the positive side, the evaluation is much more straightforward:

#bb(L_(+)) = lim_(x->0^(+)) 1/(x^2e^(-1//x))#

#= 1/(0 cdot L_(+,e^(-1//x))) = 1/(0cdot0) = 1/0 = bb(oo)#

OVERALL LIMIT

#L_(-) = 0#
#L_(+) = +oo#

The limit from either side does not match.

Thus, the limit is undefined.