Limit Help: (x - 1)/(sqrt(x) - 1) as x -> 1

[-Whiplash-]

I have the same problem as this one (which is how I found this site in the first Place).

http://www.freemathhelp.com/forum/threads/45882-Limit-Help-(x-1)-(sqrt(x)-1)-as-x-gt-1

After reading through the help (which seemed to help that guy, but I'm still confused) I don't know what to do, can someone explain it to me?

Also, can someone help me find the derivative of this using newton's Quotient (F(x) = [F(x+h)-F(x)]/h)?

F(x) = x/(x+1)

 

[JeffM]

I have the same problem as this one (which is how I found this site in the first Place).

http://www.freemathhelp.com/forum/threads/45882-Limit-Help-(x-1)-(sqrt(x)-1)-as-x-gt-1

After reading through the help (which seemed to help that guy, but I'm still confused) I don't know what to do, can someone explain it to me?

Also, can someone help me find the derivative of this using newton's Quotient (F(x) = [F(x+h)-F(x)]/h)?

F(x) = x/(x+1)

Before you start a new thread, please read READ BEFORE POSTING. You will see that (a) we greatly prefer one problem per thread, (b) we need background on what you are currently studying and your age so we can fashion an answer that is right for you, and (c) we need you to show your work so far or explain what is stopping you from starting.

I would not go back to that previous thread on your limit question because the student was so confused that the thread is INDEED hard to follow.

Question 1

\(\displaystyle \displaystyle \lim_{x \rightarrow 1}\dfrac{x - 1}{\sqrt{x} - 1} = what?\)

What we have here is a common problem of finding the limit of a function that is continuous when the function's argument is close to the limiting value, but is indeterminate at the limiting value. There are at least two methods to attack such problems, but one is to find a DIFFERENT continuous function that has the same values in the neighborhood of the limiting value and is not indeterminate there.

That is pretty abstract, but in this case what is means is that \(\displaystyle \sqrt{1} - 1 = 0 = 1 - 1\ and \dfrac{0}{0}\ is\ indeterminate.\)

What is making this indeterminate at x = 1? It is that the denominator of the fraction is zero if x = 1. So we have to find a continuous function that either has no denominator or a different denominator, but has the same value close to 1. This takes a little creativity, but there are some tricks to learn. In this one, it is to see that the numerator can be factored in terms of the denominator to give

\(\displaystyle x \ne 1\ and\ f(x) = \dfrac{x - 1}{\sqrt{x} - 1} = \dfrac{\left(\sqrt{x} + 1\right)\left(\sqrt{x} - 1\right)}{\sqrt{x} - 1}.\)

Now if you cancel like terms, you get a new continuous function that is not a fraction but gives the same result as f(x) everywhere near x = 1 and exists at x = 1. What next?

Question 2.

For reasons that I do not remember, Newton's Quotient seems like some magical and inexplicable talisman to students just starting calculus. It certainly did to me, but it is just a recipe. First calculate

F(x + h) given that \(\displaystyle F(x) = \dfrac{x}{x + 1}.\) What do get when you do that?

 

[-Whiplash-]

Before you start a new thread, please read READ BEFORE POSTING. You will see that (a) we greatly prefer one problem per thread, (b) we need background on what you are currently studying and your age so we can fashion an answer that is right for you, and (c) we need you to show your work so far or explain what is stopping you from starting.

I would not go back to that previous thread on your limit question because the student was so confused that the thread is INDEED hard to follow.

Question 1

\(\displaystyle \displaystyle \lim_{x \rightarrow 1}\dfrac{x - 1}{\sqrt{x} - 1} = what?\)

What we have here is a common problem of finding the limit of a function that is continuous when the function's argument is close to the limiting value, but is indeterminate at the limiting value. There are at least two methods to attack such problems, but one is to find a DIFFERENT continuous function that has the same values in the neighborhood of the limiting value and is not indeterminate there.

That is pretty abstract, but in this case what is means is that \(\displaystyle \sqrt{1} - 1 = 0 = 1 - 1\ and \dfrac{0}{0}\ is\ indeterminate.\)

What is making this indeterminate at x = 1? It is that the denominator of the fraction is zero if x = 1. So we have to find a continuous function that either has no denominator or a different denominator, but has the same value close to 1. This takes a little creativity, but there are some tricks to learn. In this one, it is to see that the numerator can be factored in terms of the denominator to give

\(\displaystyle x \ne 1\ and\ f(x) = \dfrac{x - 1}{\sqrt{x} - 1} = \dfrac{\left(\sqrt{x} + 1\right)\left(\sqrt{x} - 1\right)}{\sqrt{x} - 1}.\)

Now if you cancel like terms, you get a new continuous function that is not a fraction but gives the same result as f(x) everywhere near x = 1 and exists at x = 1. What next?

Thanks a lot. I skimmed over the rules before posting, but clearly I should've taken a better look. I just started Pre-Calculus a week and a half ago, and I'm doing it by correspondence, so I have no one to help me, then I found this site and it seemed like a god send, I'm sorry. But thanks to you I was able to find the answer to the problem was 2, I would've NEVER thought of doing that since it requires, as you said, "creativity".

Question 2.

For reasons that I do not remember, Newton's Quotient seems like some magical and inexplicable talisman to students just starting calculus. It certainly did to me, but it is just a recipe. First calculate

F(x + h) given that \(\displaystyle F(x) = \dfrac{x}{x + 1}.\) What do get when you do that?

I'm at this: \(\displaystyle \dfrac{\dfrac{x + h}{x + h + 1} - \dfrac{x}{x + 1}}{h}\). I don't know what I'm supposed to do next since the book doesn't have an example of question like this and I can't use the rules for this one (because the assignment ay I have to use then Newton Quotent)

 

[pka]

I'm at this: \(\displaystyle \dfrac{\dfrac{x + h}{x + h + 1} - \dfrac{x}{x + 1}}{h}\). I don't know what I'm supposed to do next since the book doesn't have an example of question like this and I can't use the rules for this one (because the assignment ay I have to use then Newton Quotent)

Multiply numerator and denominator by \(\displaystyle (x+h+1)(x+1)\).

 

[JeffM]

Thanks a lot. I skimmed over the rules before posting, but clearly I should've taken a better look. I just started Pre-Calculus a week and a half ago, and I'm doing it by correspondence, so I have no one to help me, then I found this site and it seemed like a god send, I'm sorry. But thanks to you I was able to find the answer to the problem was 2, I would've NEVER thought of doing that since it requires, as you said, "creativity".

Just remember that WE may not remember your background so you will have to remind us. Calculus is the first math course where, outside of word problems, you frequently have to just try things. It cannot be learned by rote. You learn by rote a bunch of tools, but you cannot completely learn by rote when to apply what tool. So search in your tool box (which admittedly is pretty empty after a week) and try them one after another if you are stuck.

I'm at this: \(\displaystyle \dfrac{\dfrac{x + h}{x + h + 1} - \dfrac{x}{x + 1}}{h}\). I don't know what I'm supposed to do next since the book doesn't have an example of question like this and I can't use the rules for this one (because the assignment ay I have to use then Newton Quotent)

I like to do this kind of problem in steps.

STEP 1. Write down F(x + h).

\(\displaystyle F(x) = \dfrac{x}{x + 1} \implies F(x + h) = \dfrac{x + h}{(x + h) + 1}.\)

STEP 2. Simplify F(x + h) - F(x).

\(\displaystyle F(x + h) - F(x) = \dfrac{x + h}{(x + h) + 1} - \dfrac{x}{x + 1} = \dfrac{(x + h)(x + 1)}{\{(x + h) + 1\}(x + 1)} - \dfrac{x(x + h + 1)}{\{(x + h) + 1\}(x + 1)} = \dfrac{x^2 + x + hx + h - (x^2 + hx + x)}{\{(x + h) + 1\}(x + 1)} \implies.\)

\(\displaystyle F(x + h) - F(x) = \dfrac{h}{\{(x + h) + 1\}(x + 1)}.\)

STEP 3. Divide by h or multiply by 1/h, where h is not zero, and "simplify" some more.

\(\displaystyle \dfrac{F(x + h) - F(x)}{h} = \dfrac{h}{\{(x + h) + 1\}(x + 1)} * \dfrac{1}{h} = \dfrac{1}{\{(x + h) + 1\}(x + 1)}.\)

STEP 4. Only now take the limit.

\(\displaystyle \displaystyle \lim_{h \rightarrow 0}\left(\dfrac{1}{\{(x + h) + 1\}(x + 1)}\right) = what?\)

 

[-Whiplash-]

Thank you so much. I never would have thought to multiply it by one over h.

 

[JeffM]

Thank you so much. I never would have thought to multiply it by one over h.

You're welcome but multiplying by 1/h and dividing by h are completely equivalent. Sometimes one is more intuitive to me, and sometimes the other is more intuitive to me, but they come out exactly the same.