How do you find the limit of #[e^(3x2-12x+12)] / (x^4-16)# as x approaches 2?

1 Answer
Aug 18, 2017

The limit does not exist.

Explanation:

As #xrarr2#, the exponent #3x^2-12x+12 rarr 0#.

So the numerator goes to #1#.

meanwhile the denominator goes to #0#.

The limit does not exist.

An investigation of the sign will reveal that

as #xrarr2# from the left the quotient decreases without bound:

#lim_(xrarr2^-)e^(3x^2-12x+12)/(x^4-16) = -oo# and

as #xrarr2# from the right the quotient increases without bound:

#lim_(xrarr2^+)e^(3x^2-12x+12)/(x^4-16) = oo#