Question

A function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity.
For example, the function f(x)= \frac{-3(x+2)}{x^2+4x+4} has a vertical asymptote at x=-2. For each of the following limits, enter either 'P' for positive infinity, 'N' for negative infinity, or 'D' when the limit simply does not exist.
\displaystyle{ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} = }
\displaystyle{ \lim_{x\to -2^+} \frac{-3(x+2)}{x^2+4x+4} =}
\displaystyle{ \lim_{x\to -2} \frac{-3(x+2)}{x^2+4x+4} =}

Answer 1

\[ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} \]

\[ \lim_{x\to -2^+} \frac{-3(x+2)}{x^2+4x+4} \]
\[\lim_{x\to -2} \frac{-3(x+2)}{x^2+4x+4} \]

Answer 2

these are one sided limits

Answer 3

thanks@ ash2326

Answer 4

I DONT UNDERSTAND one sided limits

Answer 5

i know the limit if undefined at -2

Answer 6

and thats abt all i know..please help

Answer 7

\[ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} = \]
Let
\[x=-2-h\ where\ h\to 0\]
so
\[ \lim_{h\to 0} \frac{-3(-2-h+2)}{(-2-h)^2+4(-2-h)+4} =\lim_{h \to 0}\frac{-3\times -h}{4+h^2+4h-8-4h+4} \]
We get
\[\lim_{h\to 0}\frac{3h}{h^2}=+\infty\]
@rukh now you try others this way

Answer 8

why did u use h?

Answer 9

Whenever we have limit like \(\lim_{x\to2^+}\)
it means x is approaching 2 from right side, which means x is just larger than 2, so
x can be written as x=2+h, where h is a very small quantity, approaching 0

Answer 10

Similarly
\[\lim_{x\to 2^-}\]
means x is approaching 2 from left side, which means x is just smaller than 2 or x=2-h
where h is a small quantity tending towards 0

Answer 11

ok i will try this and see if i get the correct answer

Answer 12

Yeah you know even if the limits both sides are same. If the two sides limits are infinity and at that point also if it's the same. Then also the limit won't exist

Answer 13

ok im trying to work it now

Answer 14

Good

Answer 15

wait how did u get positive infinity for the example u showed me...when i work it out, i get undefined

Answer 16

is undefined the same as + infinity????

Answer 17

If limit is infinity positive or negative or of the form like 0/0, 1^(infinity) or , infinity/infinity. Then the limit doesn't exist

Answer 18

ok. i see now.

Answer 19

so would u advise me...for future problems with this same form...to replace x with for example -2+h or -2-h?

Answer 20

or is there a shorter way to do this?

Answer 21

Yeah this is the way. No shortcuts:)

Answer 22

ok well thanks for your help. it was much needed...i was never told to approach the problem this way.

Answer 23

Welcome @rukh