Question

Help me with limits?

I don't have a specific problem that I have a question on, but I just don't really know how to do limits that you can't factor... Someone please please help?

Answer 1

Hmm there are a bunch of different kinds of limits that don't require factoring.
Do you mean something like this?\[\Large \lim_{x\to1}\frac{x^2+3x+1}{2x-1}\]Which requires direct substitution.


Or something like this maybe?\[\Large \lim_{x\to2^{+}} \frac{x-1}{x-2}\]

Answer 2

Something like the second one please. And also ones where x->infinity

Answer 3

The second one is a little tricky.
It doesn't really require much algebra, just some quick thinking.

So we can clearly see that there is an asymptote at x=2, right? (Because 2 causes our denominator to be zero).

So when x approaches 2, our function should be approaching \(\Large \infty\) or \(\Large-\infty\).
To determine which, we need to check the `sign` of the numerator and the denominator.

Do you understand what the little + means in the exponent position?

Answer 4

Yeah we're approaching the 2 from the right side. But I have some questions already haha

First of all, I kind of don't really know how to find vertical asymptotes by simply looking at the equation. Because not all functions with a denominator of 0 have vertical asymptotes right?

And also, is there really no way to solve non-factoring limits with algebra? Because I was watching Khan Academy earlier and he simply plugged numbers in and tried to graph

Answer 5

`Because not all functions with a denominator of 0 have vertical asymptotes right?`
Hmm I can't think of a case where this would be true.

Here's an example:\[\Large \frac{x^2+3x+2}{x^2-4}\]To find vertical asymptotes of this function, we need to find out what values of x make the denominator zero.

So we start by setting the denominator equal to zero and solve for those bad x values.\[\Large 0\quad=\quad x^2-4\]Factoring:\[\Large 0\quad=\quad (x-2)(x+2)\]Using our Zero Factor Property, we set each factor equal to zero and solve for x in each case,\[\Large 0=(x-2)\qquad\to\qquad x=2\]\[\Large 0=(x+2)\qquad\to\qquad x=-2\]

Answer 6

So for the second limit I posted, it will require some plugging in I guess.
But not much algebra beyond that.

Answer 7

So if we're approaching 2 `from the right`, that means our value will always be `slightly larger than 2`, right?

Answer 8

So let's plug in a value slightly larger than 2 to figure out the `sign` of the numerator and denominator.

Answer 9

\[\Large \lim_{x\to2^{+}} \frac{x-1}{x-2} \quad\approx\quad \frac{2.00001-1}{2.00001-2}\]

Answer 10

Um really quick, for your example with the denominator as x^2-4, I just graphed it and there is a vertical asymptote at 2 but not negative 2

Answer 11

Oh and would the limit for (x-1)/(x-2), would it be -infinity?

Answer 12

wait just kidding +infinity?

Answer 13

Ah that was a bad example I gave :( Sorry bout that..
In that example, the factors (x+2) cancel out from the numerator and denominator so we don't get an asymptote there.

The function is still undefined at x=-2, but no asymptote.
So ya I guess your points about it not always causing an asymptote is well put :)

Answer 14

So in this limit, both the numerator and denominator are giving us positive values, yes +infinity sounds right!

Answer 15

OHHHHH i get it. So if it factors out, it's probably just a removable discontinuity or jump or something but it it doesn't get canceled out, it would either be negative or positive infinity?

Answer 16

Yessss, good observation :3

Answer 17

So we need to look at an infinite limit as well?
How bout something like this maybe?\[\Large \lim_{x\to\infty}\frac{x^2-2x-1}{x^3+3}\]

Answer 18

Yes please :P

Answer 19

Again this doesn't require much algebra.
You can follow these simple rules or we can think about it more logically if you need to make sense of it.

Here are some rules:
`Degree of Denominator > Degree of Numerator`: Function is approaching zero.
`Degree of Numerator > Degree of Denominator`: Function is approaching \(\Large\infty\).
`Degree of Numerator = Degree of Denominator`: Function is approaching the ratio of the leading coefficients.

Answer 20

So to maybe make some sense of this...
See how the denominator is of a higher degree than the numerator?
It means the denominator is approaching infinity much much faster than the numerator.

Here is something to keep in mind with polynomials when we have infinite limits.
Nothing matters except the leading terms.
Everything else is approaching infinity too slowly to have any type of impact.

So we can actually think of the problem like this:\[\Large \lim_{x\to\infty}\frac{x^2}{x^3}\]

Answer 21

Oh so when you're talking about the degrees, you're only looking at the leading terms as well?

Answer 22

Ah yes I should have been more clear.
When I say degree, I mean the largest degree of x.

Answer 23

Haha yeah of course. But before you posted the post about only looking at leading terms, I was looking at the -x in the numerator too haha XD

Wow that was super super helpful! My teachers never even taught me the rules.

One last thing, in the problem with discontinuities at 2 and -2, how would I find the limit as x->-2, since it isn't simply either + or - infinity

Answer 24

In this problem?\[\Large \lim_{x\to-2}\frac{x^2+3x+2}{x^2-4}\]Let's simplify this down a little bit.\[\Large \lim_{x\to-2}\frac{(x+1)(x+2)}{(x-2)(x+2)}\]We'll cancel out the common factor but we need to place a restriction on x when we do,\[\Large \lim_{x\to-2}\frac{(x+1)}{(x-2)}, \qquad\qquad (x+2)\ne0\qquad\to\qquad x\ne-2\]

Answer 25

From here we can see that we have a removable discontinuity `at` x=-2.
But that won't affect our limits will it?
We can simply plug x=-2 directly in to see what it's approaching.

Answer 26

Example:So like if we had some function with a removable discontinuity, we just need to make sure that the limits from the left and right agree ( to ensure there is no jump ).

I think the only way we're getting a jump anyway is if the function is piecewise, so I don't think we need to worry about that.