Need help with a homework problem

[golden_ratio]

I know the answer is 1/2 from the back of the book. But I can’t reverse engineer that answer to figure out how the problem is solved.
Things I know:
• Lim h—>0 is generally referring to the limit of the function as the slope approaches 0
•g(-2) = 1 and g’(-2)=0. I don’t know what g(-2+h) is. The slope is zero, so g(-2+h) would be zero. But (0-1.5)/0 is undefined. So I’m completely wrong there.
Idk. Any help?

 

[Dr.Peterson]

I know the answer is 1/2 from the back of the book. But I can’t reverse engineer that answer to figure out how the problem is solved.
Things I know:
• Lim h—>0 is generally referring to the limit of the function as the slope approaches 0
•g(-2) = 1 and g’(-2)=0. I don’t know what g(-2+h) is. The slope is zero, so g(-2+h) would be zero. But (0-1.5)/0 is undefined. So I’m completely wrong there.
Idk. Any help?

First, h is not the slope. It is a small change in x.

Second, the expression looks a lot like the difference quotient used in defining the derivative, but it isn't the same. (What is g(-2)?) Don't let the similarity distract you.

You might be able to use some thinking related to the derivative; or you might do just what you did at the end: just observe that the numerator does not go to zero (though you're wrong that g(-2+h)=0), while the denominator does.

Maybe you are not completely wrong! You just have a couple details to correct.

 

[golden_ratio]

First, h is not the slope. It is a small change in x.

Second, the expression looks a lot like the difference quotient used in defining the derivative, but it isn't the same. (What is g(-2)?) Don't let the similarity distract you.

You might be able to use some thinking related to the derivative; or you might do just what you did at the end: just observe that the numerator does not go to zero (though you're wrong that g(-2+h)=0), while the denominator does.

Maybe you are not completely wrong! You just have a couple details to correct.

g(-2)=1
So based on what you’re saying, the equation could be restated as “the limit as a small change in x approaches 0 of (g(-2+[a small change in x])-1.5)/[a small change in x]”

since g(-2)=1, would it be “the limit as a small change in x approaches 0 of (1+[a small change in x]-1.5)/[a small change in x]” where I would plug in values approaching 0 from the left and the right for h? I’m sorry, I’m sure I’m wrong about this and sound stupid right now. It’s just not clicking for me yet.

 

[golden_ratio]

First, h is not the slope. It is a small change in x.

Second, the expression looks a lot like the difference quotient used in defining the derivative, but it isn't the same. (What is g(-2)?) Don't let the similarity distract you.

You might be able to use some thinking related to the derivative; or you might do just what you did at the end: just observe that the numerator does not go to zero (though you're wrong that g(-2+h)=0), while the denominator does.

Maybe you are not completely wrong! You just have a couple details to correct.

Here’s the work I did trying to figure it out.

 

[blamocur]

since g(-2)=1, would it be “the limit as a small change in x approaches 0 of (1+[a small change in x]-1.5)/[a small change in x]”

This does not look right: generally [imath]g(-2+x) \neq g(-2)+x[/imath]

 

[golden_ratio]

This does not look right: generally [imath]g(-2+x) \neq g(-2)+x[/imath]

Ok so I know what I’m doing wrong. Can you tell me how to correct it? Or point me in the right direction?

 

[blamocur]

I’m sorry, I’m sure I’m wrong about this and sound stupid right now.

While you are wrong you don't sound stupid at all. To me you sound much smarter then some of the posters who just dump their homework and ask for a full and detailed solution. IMHO, trying and making mistakes is the smartest way to learn.

 

[golden_ratio]

While you are wrong you don't sound stupid at all. To me you sound much smarter then some of the posters who just dump their homework and ask for a full and detailed solution. IMHO, trying and making mistakes is the smartest way to learn.

Thanks so much, I’m a math major and really do want to learn it. I posted a white board of me trying to figure this out, but the limit isn’t approaching .5. Could you take a look and let me know if I’m doing something wrong?

 

[Dr.Peterson]

g(-2)=1
So based on what you’re saying, the equation could be restated as “the limit as a small change in x approaches 0 of (g(-2+[a small change in x])-1.5)/[a small change in x]”

since g(-2)=1, would it be “the limit as a small change in x approaches 0 of (1+[a small change in x]-1.5)/[a small change in x]” where I would plug in values approaching 0 from the left and the right for h? I’m sorry, I’m sure I’m wrong about this and sound stupid right now. It’s just not clicking for me yet.

All that you're missing here is that where you have "1+[a small change in x]", you left out the fact that you are applying g to "-2+[a small change in x]". That is, you first add the small change, and then apply function g.

What is the value of g(x) when you add or subtract a small amount (say, something less than 1) to x? Put that into the formula.

Thanks so much, I’m a math major and really do want to learn it. I posted a white board of me trying to figure this out, but the limit isn’t approaching .5. Could you take a look and let me know if I’m doing something wrong?

What whiteboard? Now I see it; it wasn't visible before. You're doing the right things; what happens when you get even closer?

And why do you expect the limit to be 0.5?

 

[golden_ratio]

All that you're missing here is that where you have "1+[a small change in x]", you left out the fact that you are applying g to "-2+[a small change in x]". That is, you first add the small change, and then apply function g.

What is the value of g(x) when you add or subtract a small amount (say, something less than 1) to x? Put that into the formula.


What whiteboard?

And why do you expect the limit to be 0.5?

Photo attached. And the back of the book says .5 is the limit. But from my calculation below, I see that limit does not exist because from the left, the limit approaches infinity and from the right the limit approaches negative infinity.

 

[Dr.Peterson]

Photo attached. And the back of the book says .5 is the limit. But from my calculation below, I see that limit does not exist because from the left, the limit approaches infinity and from the right the limit approaches negative infinity.

Right. They apparently got it wrong, and I can't see what they might have been thinking. (By the way, it would have been helpful if you'd mentioned that from the start.)

The expression before taking the limit is -0.5/h, and the limit does not exist, as you say.

 

[golden_ratio]

Right. They apparently got it wrong, and I can't see what they might have been thinking. (By the way, it would have been helpful if you'd mentioned that from the start.)

The expression before taking the limit is -0.5/h, and the limit does not exist, as you say.

Oh I’m sorry about that. Thanks so much for the response.

 

[Dr.Peterson]

Right. They apparently got it wrong, and I can't see what they might have been thinking. (By the way, it would have been helpful if you'd mentioned that from the start.)

The expression before taking the limit is -0.5/h, and the limit does not exist, as you say.

I just realized that you did say that from the start! Sorry about not noticing it and saying right off that it was wrong -- sometimes I focus too much on helping people think through the work and don't actually solve it myself first.

To make up for that, here is my full work:

[imath]\displaystyle\lim_{h\to0}\frac{g(-2+h)-1.5}{h}=\lim_{h\to0}\frac{1-1.5}{h}[/imath] because [imath]\displaystyle g(-2+h)=1[/imath] for h near 0; [imath]\displaystyle\lim_{h\to0}\frac{1-1.5}{h}=\lim_{h\to0}\frac{-0.5}{h}[/imath] which does not exist.​

Trying to reverse-engineer an answer is generally not a good idea, especially when it's wrong!

 

[golden_ratio]

I just realized that you did say that from the start! Sorry about not noticing it and saying right off that it was wrong -- sometimes I focus too much on helping people think through the work and don't actually solve it myself first.

To make up for that, here is my full work:

[imath]\displaystyle\lim_{h\to0}\frac{g(-2+h)-1.5}{h}=\lim_{h\to0}\frac{1-1.5}{h}[/imath] because [imath]\displaystyle g(-2+h)=1[/imath] for h near 0; [imath]\displaystyle\lim_{h\to0}\frac{1-1.5}{h}=\lim_{h\to0}\frac{-0.5}{h}[/imath] which does not exist.​

Trying to reverse-engineer an answer is generally not a good idea, especially when it's wrong!

Yeah, I agree that it's not ideal to reverse-engineer answers. In this instance, it actually cost me a lot of time instead of saving me time. Still, thank you again for taking the time to reply. It was incredibly helpful and I now have an understanding of the concept!