What Happens to f(x) as x Approaches 2?

[nycmathguy]

Homework Statement

Determine a Limit Algebraically

Relevant Equations

Linear Expression
Quadratic Expression

Investigate A Limit

Investigate the limit of f(x) as x tends to c at the given c number.

Attachment has been deleted.

Let me see.

Let c = 2

I think I got to take the limit of f(x) as x tends to 2 from the left and right. What about as x tends to 2 (from the left and right at the same time)?

Find the limit of (x + 2) as x tends to 2 from the left side.

(2 + 2) = 4

Find the limit of x^2 as x tends to 2 from the right side.

(2)^2 = 4

LHL = RHL

Thus, the limit of f(x) as x tends to 2 is 4.

Is this right?

What about the middle section of this piecewise function? There we see f(x) is 4 if x = 2. I think we can say concerning the middle section that the limit of f(x) as x tends to 2 from the left and right at the same time is 4.

Yes?

 

[anuttarasammyak]

If f(x) is continuous at x=c then approaches from the both sides coincide to f(c).

 

[nycmathguy]

If f(x) is continuous at x=c then approaches from the both sides coincide to f(c).

I did all this work for nothing, right?

 

[anuttarasammyak]

I just wrote the last conclusion. Effort to get it is an important issue.

 

[nycmathguy]

I just wrote the last conclusion. Effort to get it is an important issue.

Yes, I try to show my part.

You said:

"If f(x) is continuous at x=c then approaches from the both sides coincide to f(c)."

Meaning? Is this the answer?

 

[Mark44]

Homework Statement:: Determine A Limit Algebraically
Relevant Equations:: Linear Expression
Quadratic Expression

Investigate A Limit

Investigate the limit of f(x) as x tends to c at the given c number.

Attachment has been deleted.

Let me see.

Let c = 2

I think I got to take the limit of f(x) as x tends to 2 from the left and right. What about as x tends to 2 (from the left and right at the same time)?

Find the limit of (x + 2) as x tends to 2 from the left side.

(2 + 2) = 4

Find the limit of x^2 as x tends to 2 from the right side.

(2)^2 = 4

LHL = RHL

Thus, the limit of f(x) as x tends to 2 is 4.

Is this right?

What about the middle section of this piecewise function? There we see f(x) is 4 if x = 2. I think we can say concerning the middle section that the limit of f(x) as x tends to 2 from the left and right at the same time is 4.

Yes?

You've written a lot of stuff here, but much of it is extraneous (e.g., "let me see" etc., but you have omitted some important parts, such as the formula for the function (or an image of it).

What middle section are you asking about? Are there three piecewise definitions for this function, something like this:
$f(x) = \begin{cases}x + 2, & x < 2 \\ 4, & x = 2 \\ x^2, & x > 2 \end{cases}$

 

[nycmathguy]

You've written a lot of stuff here, but much of it is extraneous (e.g., "let me see" etc., but you have omitted some important parts, such as the formula for the function (or an image of it).

What middle section are you asking about? Are there three piecewise definitions for this function, something like this:
$$f(x) = \begin {cases} x + 2 & x < 2 \\
4 & x = 2 \\
x^2 & x > 2 \end{cases}$$
f(x) = x + 2, for x

I deleted the picture by mistake a few days ago.

 

[nycmathguy]

You've written a lot of stuff here, but much of it is extraneous (e.g., "let me see" etc., but you have omitted some important parts, such as the formula for the function (or an image of it).

What middle section are you asking about? Are there three piecewise definitions for this function, something like this:
$$f(x) = \begin {cases} x + 2 & x < 2 \\
4 & x = 2 \\
x^2 & x > 2 \end{cases}$$
f(x) = x + 2, for x

Correct. This is the correct piecewise function.
Note: f(x) is 4 when x = 2 is what I call the middle part. I also call (x + 2) the top part and of course, x^2 is the bottom part. All three together make one function, a piecewise function.

 

[Mark44]

So if $\lim_{x \to 2^-}f(x)$ and $\lim_{x \to 2^+}f(x)$ both exist and are both equal to f(2), then $\lim_{x \to 2}f(x)$ exists and is equal to f(2) which is 4.

The notations $2^-$ and $2^+$ represent approaches from the left and right, respectively.

 

[nycmathguy]

So if $\lim_{x \to 2^-}f(x)$ and $\lim_{x \to 2^+}f(x)$ both exist and are both equal to f(2), then $\lim_{x \to 2}f(x)$ exists and is equal to f(2) which is 4.

The notations $2^-$ and $2^+$ represent approaches from the left and right, respectively.

Very good. Didn't I conclude the limit is 4?

 

[Mark44]

In post #5 you weren't sure that you had answered the question.

 

[nycmathguy]

In post #5 you weren't sure that you had answered the question.

Mark,

I am receiving lots of warnings. Why? I am posting calculus questions in the calculus and beyond forum and precalculus in the precalculus forum. Why the warnings?

 

[Mark44]

Mark,

I am receiving lots of warnings. Why? I am posting calculus questions in the calculus and beyond forum and precalculus in the precalculus forum. Why the warnings?

You currently have three warnings. One warning was for posting the URL for downloads of copyrighted books. The other two warnings were for post homework questions in forum sections other than the homework sections.

 

[nycmathguy]

You currently have three warnings. One warning was for posting the URL for downloads of copyrighted books. The other two warnings were for post homework questions in forum sections other than the homework sections.

I am starting out on the wrong track here.

 

[vela]

What about the middle section of this piecewise function? There we see f(x) is 4 if x = 2. I think we can say concerning the middle section that the limit of f(x) as x tends to 2 from the left and right at the same time is 4.

The middle piece has nothing to do with the limit. You only care about what $f$ does as $x$ approaches 2, not about what $f$ actually does at $x=2$. If f(2) was equal to, say, 0 or even if it was undefined, you would still conclude that the limit as $x \to 2$ exists and is equal to 4 because you showed both one-sided limits are equal to 4.

For the function you were given, since you have $\lim_{x \to 2} f(x) = f(2)$, you could also conclude that $f$ is continuous at $x=2$.

 

[nycmathguy]

The middle piece has nothing to do with the limit. You only care about what $f$ does as $x$ approaches 2, not about what $f$ actually does at $x=2$. If f(2) was equal to, say, 0 or even if it was undefined, you would still conclude that the limit as $x \to 2$ exists and is equal to 4 because you showed both one-sided limits are equal to 4.

For the function you were given, since you have $\lim_{x \to 2} = f(2)$, you could also conclude that $f$ is continuous at $x=2$.

Good to know. As far as continuity is concerned, this concept is coming up a few sections later. I think I am posting too many questions. This is a very bad study habit. I will post no more than 3 problems per week per textbook.