Expert: Vineeth Venugopal - 1/24/2009

Question
Hi, I was wondering how I would go about doing the following limit question.

lim       (square root of x+5)-3/
x--> 4               x-4

I can't find any good examples of how to do limits with radicals and non radicals in the numerator.


Any help would be greatly appreciated

Answer
Hai Jake,


 I don't understand why you keep this question private.

There are two ways to do this question. The first is question specific and hence applies only to this prob. The second is a general method applicable to all questions on limits.

[(x+5)^(1/2)-3]/(x-4)= [(x+5)^(1/2)-3]/(x+5-9)=
[(x+5)^(1/2)-3]/[(x+5)^(1/2)-3][(x+5)^(1/2)+3]=1/[(x+5)^(1/2)+3]

thus lim x->4 [(x+5)^(1/2)-3]/(x-4)= lim x->4 1/[(x+5)^(1/2)+3]
= 1/(3+3)=1/6;

In the second method take the derivative of both the numerator and denominator. now take the limit. The limit remains unchanged by this operation.

ie,

lim x-> a f(x)/g(x) = lim x->a f'(x)/g'(x).

thus lim x->4 [(x+5)^(1/2)-3]/(x-4)= lim x->4 1/2*(x+5)^(1/2)= 1/2*3=1/6

hope this helps