# Solve Question on Limits of f(x) at x=-2

• poli275
In summary, the conversation is about finding the limit of a function at x = -2. The function has a "hole" at x = -2, but the limit exists and is equal to 9. The conversation also discusses the concept of one-sided limits and the fact that a limit is not always a tangible value.
poli275

## Homework Statement

Hi everyone I just have this question in my exam, it's a limit question and I don't know how to solve it. Any help are much appreciated :D

## Homework Equations

http://img42.imageshack.us/img42/8578/47934321.jpg

Find Lim (x -> -2) f(x) if it exist

(sorry I don't know how to type the code)

## The Attempt at a Solution

The answer I gave was the limit does not exist. I din't give an explanation for that.

Thanks again for any help :D

Last edited by a moderator:
Look at the one sided limits as x --> -2 from the left and right.

Ok I looked at the graph, and both the curve and line meet at x = -2 and y value is 9, do does that means that the limit of the function exist and it's at y = 9?

poli275 said:
Ok I looked at the graph, and both the curve and line meet at x = -2 and y value is 9, do does that means that the limit of the function exist and it's at y = 9?

Yes. If the right and left limits at a point exist and are equal, then their common value is the limit of the function. On an exam you would likely be expected to give some justification for your conclusion that the right and left limits are both 9.

Ok thanks for your help. I jhave one more question on the question itself. Because for the x^2 + 5, it was given that x is less than x is less than -2, what does that actually means? Thanks.

As LCKurtz said initially, look at the "right" and "left" limits. To the left of x= -2, the function is just $x^2+ 5$. That is a polynomial and so continuous for all x. In particular it is continuous at x= -2 and so its limit, as x approaches -2, is the value of the function there, $(-2)^3+ 5= 9$.
[tex]\lim_{x\to -2} X^2+ 5= 9[/itex].
[tex]\lim_{x\to -2^-} f(x)= 9[/itex].

To the right of x= -2, the function is 3- 3x. Again, that is a polynomial. It is also continuous at x= -2 and so
$\lim_{x\to -2} 3- 3x= 3- 3(-2)= 9$
$\lim_{x\to -2^+} f(x)= 9$.

Since the two one-sided limits exist and are equal, the limit itself exists and is that common value, 9.

Ok thanks for the help, I will try to figure it out the whole thing again. Thanks again.

I find it helpful to remember that a limit is not necessarily a tangible thing. As you saw in the function you were given there is a "hole" in the graph so -2 doesn't really exist, but the limit is in fact there. When you are looking for the limit, you are just looking for where the function would be if it actually existed at that point, whether it actually does exist there or not.

Good luck!

## 1. What is the limit of f(x) at x=-2?

The limit of f(x) at x=-2 is the value that f(x) approaches as x gets closer and closer to -2 from both sides.

## 2. How do you solve for the limit of f(x) at x=-2?

To solve for the limit of f(x) at x=-2, you can use the direct substitution method or evaluate the left and right-hand limits and then compare them. If they are equal, that is the limit. If they are different, the limit does not exist.

## 3. Can the limit of f(x) at x=-2 be a different value from f(-2)?

Yes, the limit of f(x) at x=-2 and the value of f(-2) can be different. The limit is based on the behavior of f(x) as x approaches -2, while the value of f(-2) is just the output when x is exactly equal to -2.

## 4. Is it possible for the limit of f(x) at x=-2 to not exist?

Yes, the limit of f(x) at x=-2 may not exist if the left and right-hand limits are not equal or if they approach different values. This could happen if there is a vertical asymptote or a jump discontinuity at x=-2.

## 5. Can you use any method to solve for the limit of f(x) at x=-2?

There are several methods that can be used to solve for the limit of f(x) at x=-2, such as direct substitution, factoring and simplifying, using algebraic rules, or graphing. The method used may depend on the complexity of the function and the tools available.

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