Understanding Continuity and Limits for Homework Success

[Frankie715]

Homework Statement

1)

2)

Homework Equations

The Attempt at a Solution

1) I have done plenty of these, but this one is stumping me. I tried plugged in 0 approach for h and I got 11-11=0. With h on the bottom as 0. I know this isn't the right answer. I also know the limit does exist. If anyone could help me with HOW you come to the conclusion I would greatly appreciate it.

2) I have pasted the problem with my attempts at an answer. I am having a little bit of trouble here trying to determine exactly what intervals are continuous and why.Thanks guys for any help you can give!

 

[DrummingAtom]

Try using an Algebraic Conjugate.

 

[eumyang]

1) Rationalize the NUMERATOR.

2) By my count, there are two more intervals where g is continuous. What is happening at x = 4?

 

[Frankie715]

1) Rationalize the NUMERATOR.

2) By my count, there are two more intervals where g is continuous. What is happening at x = 4?69

1) I got 11 out of the square root of 121 which cancels with the other -11. Confused about the remaining h.

2) I was confused at x = 4. Are my current 3 answers correct?

 

[eumyang]

1) No, you can't do that. You have a square root of a SUM, and you can't split into a sum of two square roots:
$$\sqrt{121 + h} \ne 11 + \sqrt{h}$$

You rationalize the numerator by multiplying top and bottom by the numerator's conjugate. For example, if the numerator of a fraction is
$$a + \sqrt{b}$$,
then you multiply top and bottom by
$$a - \sqrt{b}$$,
because in doing so, you make the numerator a rational number (hence, rationalizing).
$$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$.

 

[Frankie715]

Ok, let me figure this out.

So I multiply the numerator and the denominator by

sqrt (121+h)+11?

 

[eumyang]

Yes. Keep going. What happens next?

 

[n.karthick]

If you substitute h=0 you get 0/0 ie., an indeterminate form. So you can use L' Hospital's rule for this problem.
i.e., differentiate numerator and denominator separately and find the limit of their ratios.

 

[Frankie715]

Yes. Keep going. What happens next?

I got h on the top. (sqrt (h + 121)) h + 11h on bottom. Is that correct?

 

[Bohrok]

I got h on the top. (sqrt (h + 121)) h + 11h on bottom. Is that correct?

That's right. Now notice that each term in the denominator has a factor of h. After you factor out the h, what can you do next?

 

[eumyang]

If you substitute h=0 you get 0/0 ie., an indeterminate form. So you can use L' Hospital's rule for this problem.
i.e., differentiate numerator and denominator separately and find the limit of their ratios.

This may not be helpful if the OP is beginning Calc. I at a uni. in the U.S. He/she may have just started to study limits and so he/she would not know what L'Hôpital's rule (not L'Hospital ) is.