Solving Simple Limits Questions with No Calculator

[gamma-ray-burst]

hey hi.

i'm just starting differentials and now we are seeing the limits.
i have a question.

there is one example that goes like this :

lim 4 / (x²+2x+1)
x--> -1

well with simple calculations using -1(-) and -1(+), i know that the answer is infinity.
plus, if i trace the graph, there's an asymptote at -1.
but I'm not supposed to use a calculator.
well at the exam we can't have one.

so i do it by hand.
when i plug, let's say, -1.000001 and -0.999999, it gives me 4/0 for both.
can i say that it's infinity, or would it have to be 4/0(+)?

i feel that it's really easy, yet I'm stuck :(

also, if i get something like (1(+))², does it stays 1(+) or does it becomes like 1.0000000002?

thanks!

 

[tiny-tim]

Hi gamma-ray-burst!

lim 4 / (x²+2x+1)
x--> -1

well with simple calculations using -1(-) and -1(+), i know that the answer is infinity.
plus, if i trace the graph, there's an asymptote at -1.
but I'm not supposed to use a calculator.
well at the exam we can't have one.

so i do it by hand …

hmm … rather long-winded …

rewrite it as a lim_{ε -> 0}, where x = -1 + ε.

 

[CompuChip]

In this case, you can "plug in" x = -1. Then you find that you get something of the form 4 / 0. Therefore, the limit does not exist; we sometimes write
\lim_{x \to -1} \frac{4}{x^2 + 2x + 1} = \infty
to indicate this (\infty being the symbol for infinity -- note that this is like a convention to say that the limit does not exist, it does not mean that the limit is "equal to infinity" or something like that).

In principle, this method works: when you get something of the form number/0 or infinity/number (where in both cases, number is not equal to 0), the limit does not exist. Again, I should point out that "of the form" is a necessary addition here: technically you are not allowed to write "4/0", for example, that is an undefined expression. Only when you get something like 0/0 or infinity/infinity you really need to be careful and apply another method.

 

[HallsofIvy]

Having noted that, at x= -1, the numerator is 4 and the denominator is 0, your answer should be that the limit does not exist.

Strictly speaking,
\lim_{x\to -1} \frac{4}{x^2+ 2x+ 1}= \infty
is not correct because
\lim_{x\to -1^+} \frac{4}{x^2+ 2x+ 1}= \infty
while
\lim_{x\to -1^-} \frac{4}{x^2+ 2x+ 1}= -\infty

Best answer: "The limit does not exist."

 

[gamma-ray-burst]

Oh wow you guys are amazing.
Thanks a lot for all the answers!
Yet again, I was complicating things for nothing.
I know that next week we will see indeterminations of forms like "0/0" and "inf/inf" and things like that and I was afraid that "#/0" was some special case where I had to do other things.

So thanks again!

 

[l'Hôpital]

Having noted that, at x= -1, the numerator is 4 and the denominator is 0, your answer should be that the limit does not exist.

Strictly speaking,
\lim_{x\to -1} \frac{4}{x^2+ 2x+ 1}= \infty
is not correct because
\lim_{x\to -1^+} \frac{4}{x^2+ 2x+ 1}= \infty
while
\lim_{x\to -1^-} \frac{4}{x^2+ 2x+ 1}= -\infty

Best answer: "The limit does not exist."

This is untrue.

{x^2+ 2x+ 1} = (x+1)^2.

Whatever side you approach, the square will make it positive, thus positive infinity.