finding limits?

Question

i am sorta lost on finding limits, i think i understand the concepts, but am lacking in algebra possibly.

here is a prime question:

evaluate the limit lim x->7 (sqrt(x+2)+3/(x-7)

every time i try the problem, i end up with zero on the top and zero on the bottom... and i don't see anything that can factor.
what am i missing?

so how do i know what the conjugate is?

ok here is the full problem

evaluate .... limit as x→7 (sqrt(x+2)-3)/(x-7)

Answer 1

conjugate of (a-b) is (a+b) and vice versa

introduce the conjugate { sqrt(x+2) + 3 }
in numerator and denominator
simplify using identity (a+b)(a-b) = a^2 - b^2
cancel off (x-7)
then
apply limit


or use L-Hospital's rule

Answer 2

It seems you have made an error in the question.
It should have been
lim x->7 [√(x+2) - 3] / (x - 7)
and not
lim x->7 [√(x+2) + 3] / (x - 7)

a + √b and a - √b are conjugate of each other. In this problem, we multiply numerator and denominator with conjugate of [√(x+2) - 3] which is [√(x+2) + 3]. Thus,

lim x->7 [√(x+2) - 3] / (x - 7)
= lim x->7 [√(x+2) - 3] *[√(x+2) + 3] / (x - 7)*[√(x+2) + 3]
= lim x->7 [ x + 2 - 9 ] / (x - 7)*[√(x+2) + 3]
= lim x->7 (x - 7) / (x - 7)*[√(x+2) + 3]
= lim x->7 1 / [√(x+2) + 3]
= 1/6.

Answer 3

Edit: Okay, just saw your restated problem. This one is a 0/0 form. The way we handle this is by using the following identity:

(a+b)(a-b) = a²-b²

In this case, we have an expression of the form (a-b) in the numerator, so by multiplying both numerator and denominator by a+b (which is called the conjugate of a-b), we will get rid of the annoying square root in the numerator, and get something we can cancel. Observe:

[x→7]lim (√(x+2) - 3)/(x-7)
[x→7]lim (√(x+2) - 3)(√(x+2) + 3)/((x-7)(√(x+2) + 3))
[x→7]lim (x+2 - 9)/((x-7)(√(x+2) + 3))
[x→7]lim (x-7)/((x-7)(√(x+2) + 3))
[x→7]lim 1/(√(x+2) + 3)
1/(√(7+2) + 3)
1/(√9+3)
1/6

And we are done.

Answer 4

it would be helpful if you used more brackets to let us know what the eqn is