Question

Help- find the limit of:

Answer 1

\[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ x^2+2x-8 } \right)+\left(\frac{ x-6 }{ x^2-4} \right)\]

Answer 2

so I found that \[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ (x+4)(x-2) } \right)+\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\]
Now what?????

Answer 3

now subsitute 2 in place of x

Answer 4

it will be 6/0. error. I can't do that yet.

Answer 5

1/0 = infinity

Answer 6

sorry 1/3.
I see the next step in the book but I don't understand how they got to it. I will write it up.

Answer 7

\[\lim_{x \rightarrow 2}\left( \frac{ 6(x+2)+(x-6)(x+4) }{ (x+4)(x-2)(x+2) } \right)\]

Answer 8

they took the LCM in this step

Answer 9

ok. Can you explain that part to me? I don't seem to get it.

Answer 10

write all the terms in denominator if some term is repeating more then once , just write it one time. like in this case on one side u have (x+4)(x−2) and on other side (x+2)(x−2)
so write them all in denominator but (x-2) is written twice so u will just write it one time

Answer 11

ok but what about the numerator? How did they get to that part?

Answer 12

ok now move back a step and see see what u had in denominator before . u had (x+4)(x−2) right?

Answer 13

so you compare the old denominator with new numerator whichever is the common denominator u cancel them and whatever is left u multiply it with numerator