Calculus Limits?

Question

Can you show me the steps of how to do the limit as x approaches 2 of [sqrt (x^2+6x)-4]/x-2? Thanks. Also can be written:

(X^2+6x)^.5 -4

over

x-2.

Answer 1

You have to use the conjugate.

So Sqrt[x^2 + 6x] - 4 * Sqrt[x^2 +6x] + 4

Will result in x^2 + 6x -16 all over (x-2) * Sqrt[x^2 +6x] + 4

Factor and you get
(x-2)(8+x) all over (x-2)

then cancel out x-2 and then replace x with 2

and the limit is a 10

Answer 2

Rationalize the numerator by multiplying the expression by (sqrt(x^2+6x) plus 4)/(sqrt(x^2+6x) plus 4). The numerator will be factorable and you can cancel the factor of (x-2). Then you can evaluate the limit as usual...

Answer 3

♠ thus y=(√(x^2 +6x) –4)/(x-2);
y = {(√(x^2 +6x) –4)/(x-2)} *
* {(√(x^2 +6x) +4) / (√(x^2 +6x) +4)} =
= {(x^2 +6x –4^2)/(x-2)} / (√(x^2 +6x) +4) =
= {(x-2)*(x+8)/(x-2)} / (√(x^2 +6x) +4)=
= (x+8)/(√(x^2 +6x) +4);
♣ y(2)= (2+8)/(√(2*2 +6*2) +4) = 5/4;