Question

Calculate each of the following limits:
1.) lim(x^2+6x-5)
x->2

2.) lim(x^2-16)/(x+4)
x->-4

Answer 1

factor, cancel, replace \(x\) by \(-4\) for the second one

Answer 2

first one plug in the number, see what you get

Answer 3

i.e. compute
\[2^2+6\times 2-5\]

Answer 4

So then the first one would 11

Answer 5

The first one can be done with plain substituion, i.e. just plug in 2 and evaluate.

The second one requires some factoring and cancellation like this:\[\bf \lim_{x \rightarrow -4}\frac{ x^2-16 }{ x+4 }= \lim_{x \rightarrow -4}\frac{ (x-4)\cancel{(x+4)} }{\cancel{ x+4} }\lim_{x \rightarrow -4}= x-4\]

Answer 6

yes

Answer 7

The equal sign before x-4 isn't supposed to be there lol.

Answer 8

And substitute once again.

Answer 9

okay. I see it, so its...simple, once again.