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

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Answer 1 · 2017-08-18T21:43:26

The limit does not exist.

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#