Question about factoring for calculus

[Checkfate]

Hello, I have been reading these forums for quite a while now and am very impressed! I have a question :) I need to find the lim(x->4) (k^2-16)/(sqrt(k)-2) How would I go about factoring k^2-16 so that I can get sqrt(k)-2 on the numerator? Thanks!

 

[neutrino]

\left(\sqrt{k}\right)^4 = k^2 and 2^4 = 16.

 

[fourier jr]

i would multiply the fraction by 1, where 1 = (sqrt(x) + 2)/(sqrt(x) + 2)

i guess that's the 1st step & the only tricky part. isn't this in your textbook?

 

[Checkfate]

Thankyou both. No unfortunately it is not in my textbook. I am using my father's calculus book from university as a supplement to my regular calculus material. It was probably assumed that I knew how to solve it so it wasn't even mentioned. Unfortunately I forgot how to solve it! Anyways, thanks.

 

[fourier jr]

something you might have noticed is that it would be a very BAD idea to multiply the sqrt(x)+2 into the numerator, and probably not a good idea in general to do that. factor the x^4 - 16 & only multiply the denominators to get what you want to cancel.

 

[HallsofIvy]

At some point prior to taking calculus, you should have learned the general formula (a²- b²)= (a- b)(a+ b).