Need some help with limits and continuity

[dustinm]

I have 2 questions in regards to continuity and limits.

Question 1:
f(x)= e^{-x^{2}} if x ≠ 0.
f(x)= c if x=0.

For which value of c is f(x) continuous at x=0?

I was thinking the answer would be 1 but I feel that's incorrect.

Question 2:
Compute lim x→∞f(x).

I'm not familiar with how to solve limits to infinity when the variable is in the exponent.
Any and all help is appreciated! thank you guys.

 

[gb7nash]

I have 2 questions in regards to continuity and limits.

Question 1:
f(x)= e^{-x^{2}} if x ≠ 0.
f(x)= c if x=0.

For which value of c is f(x) continuous at x=0?

I was thinking the answer would be 1 but I feel that's incorrect.

Ok, what does it mean to be continuous at x = 0? Look at your definitions.

Question 2:
Compute lim x→∞f(x).

What does the exponent of e approach as x goes to ∞?

 

[dustinm]

Ok, what does it mean to be continuous at x = 0? Look at your definitions.

It means that f(x)=f(c) as x→c. So basically that the function can be drawn without having to lift the pen to complete the graph.

What does the exponent of e approach as x goes to ∞?

So, e^{-∞^{2}} would be approaching -∞?

 

[gb7nash]

It means that f(x)=f(c) as x→c. So basically that the function can be drawn without having to lift the pen to complete the graph.

Almost. f(x)→f(c) as x→c. In other words:

\lim_{x \to c}f(x) = f(c)

So we need to look at the left side and the right side of the equation. Replacing c with 0, what is \lim_{x \to c}f(x)? What is f(c)? Are they equal? If so...

So, e^{-∞^{2}} would be approaching -∞?

You're not thinking this all the way through. The exponent is approaching -∞ like you said. What's e raised to a large negative number?

 

[dustinm]

Almost. f(x)→f(c) as x→c. In other words:

\lim_{x \to c}f(x) = f(c)

So we need to look at the left side and the right side of the equation. Replacing c with 0, what is \lim_{x \to c}f(x)? What is f(c)? Are they equal? If so...

f(c)=0 and if you were to replace that into the equation for f(x) you would get f(x)=e^{-0^{2}} which would end up equaling 1, right?

Sorry about this, all of this is really new to me so it's tough to grasp at first.

You're not thinking this all the way through. The exponent is approaching -∞ like you said. What's e raised to a large negative number?

Ahh so it would be 0 because e^{-∞} is extremely small.

 

[gb7nash]

if you were to replace that into the equation for f(x) you would get f(0)=e^{-0^{2}} which would end up equaling 1, right?

Correct (I fixed a typo of yours). That's the right side of the equation. Now you need to look at the left side, which is:

\lim_{x \to 0}e^{-x^2}

Is this also equal to 1?

Ahh so it would be 0 because e^{-∞} is extremely small.

Correct.

 

[dustinm]

Correct (I fixed a typo of yours). That's the right side of the equation. Now you need to look at the left side, which is:

\lim_{x \to 0}e^{-x^2}

Is this also equal to 1?

Yes that would be 1.
So the final answer for making the graph continuous at 0 needs to be 1?

 

[gb7nash]

Yes that would be 1.
So the final answer for making the graph continuous at 0 needs to be 1?

Correct.

 

[dustinm]

Correct.

Thank you very much for the help with these questions!
Walking me through it helped out a bunch!