Question

Find the value of the limits: Lim x-->2 radical(x^2+6x)-4/ (x-2)?? I have an exam tonight and i need to understand this :/

Answer 1

\[\lim_{x \rightarrow 2} \frac{\sqrt{x^2+6x}-4}{x-2}\]
At x=2, the top and bottom both go to zero. That means we get to use l'hopital's rule, yay!

\[\lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\]

Answer 2

Have you heard of the L'Hôpital's rule? And are you familiar with differentiation?

Answer 3

lim_(x->2_) (sqrt(x^2+6 x)-4/(x-2)) = infinity
lim_(x->2+) (sqrt(x^2+6 x)-4/(x-2)) = -infinity
lim 2_ != lim 2+ therefore lim does not exist

Answer 4

ybarrap - You are SO wrong here bro

Answer 5

Ive seen l'hopitals rule and its confusing..

Answer 6

You need to differentiate both the numerator, and the denominator separately, then find the value at x = 2.

Answer 7

Exactly like Dalvoron said. Differentiate them separately (helps in NOT making a mistake) and then put them back together and find the value at x = 2