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ratelena [41]
3 years ago
11

Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→2 x2 + 4x −

12 x − 2
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0
\displaystyle\lim_{x\to2}\frac{x^2+4x-12}{x-2}

Substituting x=2 directly yields \dfrac{4+8-12}{2-2}=\dfrac00. Applying L'Hopital's rule, we have

\displaystyle\lim_{x\to2}\frac{2x+4}1=4+4=8

But note that x-2 is a factor of the numerator, which follows from the remainder theorem: if x=2, then 2^2+4(2)-12=0 which means x-2 is a factor of the numerator.

We have

x^2+4x-12=(x-2)(x+6)

and so

\displaystyle\lim_{x\to2}\frac{x^2+4x-12}{x-2}=\lim_{x\to2}\frac{(x-2)(x+6)}{x-2}=\lim_{x\to2}(x+6)=2+6=8
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