anonymous
  • anonymous
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} =}
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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ash2326
  • ash2326
\[ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} \] \[ \lim_{x\to -2^+} \frac{-3(x+2)}{x^2+4x+4} \] \[\lim_{x\to -2} \frac{-3(x+2)}{x^2+4x+4} \]
Shayaan_Mustafa
  • Shayaan_Mustafa
these are one sided limits
anonymous
  • anonymous
thanks@ ash2326

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anonymous
  • anonymous
I DONT UNDERSTAND one sided limits
anonymous
  • anonymous
i know the limit if undefined at -2
anonymous
  • anonymous
and thats abt all i know..please help
ash2326
  • ash2326
\[ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} = \] 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
anonymous
  • anonymous
why did u use h?
ash2326
  • ash2326
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
ash2326
  • ash2326
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
anonymous
  • anonymous
ok i will try this and see if i get the correct answer
ash2326
  • ash2326
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
anonymous
  • anonymous
ok im trying to work it now
ash2326
  • ash2326
Good
anonymous
  • anonymous
wait how did u get positive infinity for the example u showed me...when i work it out, i get undefined
anonymous
  • anonymous
is undefined the same as + infinity????
ash2326
  • ash2326
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
anonymous
  • anonymous
ok. i see now.
anonymous
  • anonymous
so would u advise me...for future problems with this same form...to replace x with for example -2+h or -2-h?
anonymous
  • anonymous
or is there a shorter way to do this?
ash2326
  • ash2326
Yeah this is the way. No shortcuts:)
anonymous
  • anonymous
ok well thanks for your help. it was much needed...i was never told to approach the problem this way.
ash2326
  • ash2326
Welcome @rukh

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