Question

is the answer to the question in the comments infinity?

Answer 1

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

Answer 2

HI!!

Answer 3

what happens when you plug in one?

Answer 4

oh i see, you get \(\frac{-8}{0}\) go no further

Answer 5

oh yeah i see you are allowed to say "infinity" but be careful here
the numerator is negative, how about the denominator?

Answer 6

I actually did something different

Answer 7

x is close to one but not one that means that the closer i get to one, the smaller the denominator gets hence a larger -ve fraction..so will the answer be -ve infinity because the value will keep decreasing

Answer 8

@misty1212

Answer 9

the numerator is negative, so it depends on whether the denominator is negative or positive for values of x > 1

Answer 10

that is because \(\lim_{x\to 1^+}\) means \(x>1\)

Answer 11

in any case you are right , it is \(-\infty\)

Answer 12

\[\lim_{x \rightarrow 1+}\frac{ x^2-9 }{ x^2+2x-3 }=\lim_{x \rightarrow 1+}\frac{ \left( x+3 \right)\left( x-3 \right) }{ x^2+3x-x-3 }\]
\[=\lim_{x \rightarrow 1+}\frac{ \left( x+3 \right)\left( x-3 \right) }{ x \left( x+3 \right)-1\left( x+3 \right) }\]
\[=\lim_{x \rightarrow 1+}\frac{ \left( x+3 \right)\left( x-3 \right) }{ \left( x+3 \right)\left( x-1 \right) }\]
\[=\lim_{x \rightarrow 1+}\frac{ x-3 }{ x-1 }\]
put x=1+h,h>0
\[h \rightarrow 0 ~as~x \rightarrow 1+\]
\[=\lim_{h \rightarrow 0}\frac{ 1+h-3 }{ h }=\lim_{h \rightarrow 0}\frac{ -2+h }{ h }\]
\[=\frac{ -2+0 }{ 0 }=-\infty \]

Answer 13

thank you @surjithayer