# Trying to find limit involving continuity concept

• Torshi
That's what you do when you have a piecewise-defined function like this. You have two separate functions, one for x <= c and one for x > c, and you're trying to find the value of c that makes the two functions match up at the boundary.So, as I was saying, you need to set the two limits equal to each other:√-3 IMO: It's more important to know why a method works than to know a method without understanding it.So, here are some questions:What is necessary in order that f(x) be continuous at x = c ?What is \displaystyle \lim_{x\to c^-}\,x^2-9\ ?What is \display
Torshi

## Homework Statement

Find c such that the function f(x) { x^2-9 while x≤ c and 6x-18 x > c } is continuous everywhere.

## Homework Equations

Given above. Basic algebra.

## The Attempt at a Solution

I made a number line. Showing that x^2-9 is approaching from the left side and 6x-18 is approaching from the right side given the designation of the inequality.

If there were numbers, I could easily do it, but the "c" is throwing me off.

An attempt to a similar problem with numbers would simply be chug and plug and knowing which side it comes from left or right, and if there was a variable "a" in there as well you can substitute.

I have no idea how to start the problem, would it become c^2-9 and 6c-18? and try to solve for c?

##f(x)=\begin{cases} x^2-9 & \mbox{if } x \leq c\\ 6x-18 & \mbox{if } x>c\end{cases}##

For ##f(x)## to be continuous at ##c##, you must satisfy
##\displaystyle\lim_{x\to c^{+}} f(x) = \displaystyle\lim_{x\to c^{-}} f(x)##

That is,
##\displaystyle\lim_{x\to c^{+}} x^2-9 = \displaystyle\lim_{x\to c^{-}} 6x-18##

That's basically what (it seems) you've done already, but this is the reasoning behind the step you've taken.

Torshi said:

## Homework Statement

Find c such that the function f(x) { x^2-9 while x≤ c and 6x-18 x > c } is continuous everywhere.

## Homework Equations

Given above. Basic algebra.

## The Attempt at a Solution

I made a number line. Showing that x^2-9 is approaching from the left side and 6x-18 is approaching from the right side given the designation of the inequality.

If there were numbers, I could easily do it, but the "c" is throwing me off.

An attempt to a similar problem with numbers would simply be chug and plug and knowing which side it comes from left or right, and if there was a variable "a" in there as well you can substitute.

I have no idea how to start the problem, would it become c^2-9 and 6c-18? and try to solve for c?
IMO: It's more important to know why a method works than to know a method without understanding it.

So, here are some questions:

What is necessary in order that f(x) be continuous at x = c ?

What is $\displaystyle \lim_{x\to c^-}\,x^2-9\ ?$

What is $\displaystyle \lim_{x\to c^+}\,6x-18\ ?$

SithsNGiggles said:
##f(x)=\begin{cases} x^2-9 & \mbox{if } x \leq c\\ 6x-18 & \mbox{if } x>c\end{cases}##

For ##f(x)## to be continuous at ##c##, you must satisfy
##\displaystyle\lim_{x\to c^{+}} f(x) = \displaystyle\lim_{x\to c^{-}} f(x)##

That is,
##\displaystyle\lim_{x\to c^{+}} x^2-9 = \displaystyle\lim_{x\to c^{-}} 6x-18##

That's basically what (it seems) you've done already, but this is the reasoning behind the step you've taken.

But I don't know how to move on from there.. I end up getting something like x/x = √-3

SammyS said:
IMO: It's more important to know why a method works than to know a method without understanding it.

So, here are some questions:

What is necessary in order that f(x) be continuous at x = c ?

What is $\displaystyle \lim_{x\to c^-}\,x^2-9\ ?$

What is $\displaystyle \lim_{x\to c^+}\,6x-18\ ?$

do you have to set the two limits equal?

Torshi said:
But I don't know how to move on from there.. I end up getting something like x/x = √-3
If you solve c2-9 =6c-18 for c,

how do you get x/x = √-3 ?

There is no x in c2-9 =6c-18 , and x/x = 1 .

Torshi said:
do you have to set the two limits equal?
Yes, which should be clear if you answer the question about the function being continuous at x = c .

SammyS said:
If you solve c2-9 =6c-18 for c,

how do you get x/x = √-3 ?

There is no x in c2-9 =6c-18 , and x/x = 1 .

I put x because the other poster. But, essentially wouldn't it be:

c^2-9 = 6c-18
c^2=6c-9
c^2/c = -3
..idk what to do ... if c was a number in the inequality then i could easily do this problem

alright hold on..

bump

nvm got it.

Since f is continuous everywhere, it is continuous at x = c.
So the left-hand limit must equal the right-hand limit. Thus:
c^2 - 9 = 6c - 18
c^2 - 6c + 9 = 0
(c - 3)^2 = 0
c = 3

thank you!

Last edited:
Torshi said:
do you have to set the two limits equal?

Yes, that's what I was suggesting.

## 1. What is a limit in the context of continuity?

A limit is a fundamental concept in calculus that describes the behavior of a function as the input values get closer and closer to a particular value. In the context of continuity, a limit represents the value that a function approaches as the input value approaches a specific value.

## 2. How do you find the limit of a function involving continuity?

To find the limit of a function involving continuity, you can use several methods such as direct substitution, factoring, and algebraic manipulation. However, the most reliable method is to use the rules of continuity, which state that the limit of a function is equal to the value of the function at that point.

## 3. Can a function have a limit if it is not continuous?

No, a function cannot have a limit if it is not continuous. In order for a function to have a limit at a specific point, it must be continuous at that point. This means that the function must be defined and have the same value from both the left and right sides of the point. If a function is not continuous, then the limit does not exist.

## 4. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of a function as the input values approach a particular point from one side (either from the left or the right). A two-sided limit, on the other hand, considers the behavior of the function as the input values approach the point from both sides. In order for a two-sided limit to exist, the one-sided limits from both sides must be equal.

## 5. How is continuity related to differentiability?

Continuity and differentiability are closely related concepts in calculus. A function is differentiable at a point if it is continuous at that point and has a well-defined derivative. This means that the function has a unique slope at that point, which is equal to the limit of the function's slope as the input values approach the point. In other words, continuity is a necessary condition for differentiability.

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