Question

How would I find the limit of (x^2 -9 )/(x^2 + 2x -3) as x approaches 1+ ? Better formatted in Question and efforts tried too.

Answer 1

do they have any common factors?

Answer 2

\[\lim_{x \rightarrow 1^{+}} \frac{ x ^{2} - 9 }{ x ^{2} + 2x -3 }\]
I tried that and I cancele d out (x+3) , but that left (x-1) in the denominator

Answer 3

yeah so we are left with:

(x-3)
-----
(x-1)


i would suggest a number line


<-------|-------------------------->
1

at x=3 we are at zero, thats simple enough to determine


0
<-------|------------|-------------->
1 3


anything bigger than 3 remains positive

0 +++
<-------|------------|-------------->
1 3

anything less then 3, as it approches 1 is negative

--- 0 +++
<-------|------------|-------------->
1 3

the question remains, is it going to - infinty? well yeah since we couldnt exclude it

Answer 4

so it is approaching negative infinity, because it will never get to 1?

Answer 5

\[\lim_{x \rightarrow 1+} \frac{x^2-9}{x^2+2x-3}=\lim_{x \rightarrow 1+} \frac{(x+3)(x-3)}{(x-1)(x+3)}=\lim_{x \rightarrow 1+} \frac{x-3}{x-1}\ = -\infty\]

Answer 6

correct, the closer it get to one, as it comes from the right ... will get bigger and bigger in the negative direction

Answer 7

okay thank you.

Answer 8

youre welcome