Can someone me simplify this expression....

[MrDickinson]

lim_(h->0^-) (e^(x+h)/((x+h)^2-1)-e^(x+h)/(x^2-1))/h = -(2 e^x x)/(x^2-1)^2

I know how to differentiate the expression using the quotient rule; however, I want to use the limit definition of a derivative to practice it more.This desire to practice led me into a trap! Now I just can't simplify the expression...

PLEASE give a thorough, step by step explanation. Thanks!

Here is a link to wolfram with the function

https://www.wolframalpha.com/input/...or"}&rawformassumption={"MC",""}->{"Formula"}

 

[Amentia]

Hello, I don't understand why in the second term you have e^(x+h) instead of e^x if you are using [f(x+h) - f(x)]/h.

 

[MrDickinson]

Hello, I don't understand why in the second term you have e^(x+h) instead of e^x if you are using [f(x+h) - f(x)]/h.

Sorry, I just took that from wolfram. It was supposed to be what I had typed in (didn't want to retype), but I see the problem. It seems I did not correctly type in what I wanted to take the limit of...I actually didn't even notice that because wolfram wasn't that helpful (it just says simplify without giving the steps to simplify) so I just moved on..

Here, I will upload a picture of my work to make it easier...

Also, my work is partial, but it doesn't matter because no matter what I did, I could not get the function out of an undefined form using algebra.

 

[Amentia]

I don't see very well all the lines but I think the last line is correct. You just need to find a way to continue. You have:

$$\frac{e^x}{D} \left[ \frac{e^h}{h}x^2- \frac{e^h}{h} - \frac{x^2}{h} -2x - h - \frac{1}{h} \right]$$

where D is your denominator without the h that I moved upwards. So from there you have to group term to remove the problems with infinity that you could have with terms like 1/h.

For example:

$$\frac{e^x}{D} \left[ \left(\frac{e^h}{h}- \frac{1}{h}\right)x^2 +\left(\frac{1}{h}- \frac{e^h}{h}\right) -2x - h \right]$$

And now if you know how to develop the exponential, it should be straightforward.

 

[MrDickinson]

I don't see very well all the lines but I think the last line is correct. You just need to find a way to continue. You have:

$$\frac{e^x}{D} \left[ \frac{e^h}{h}x^2- \frac{e^h}{h} - \frac{x^2}{h} -2x - h - \frac{1}{h} \right]$$

where D is your denominator without the h that I moved upwards. So from there you have to group term to remove the problems with infinity that you could have with terms like 1/h.

For example:

$$\frac{e^x}{D} \left[ \left(\frac{e^h}{h}- \frac{1}{h}\right)x^2 +\left(\frac{1}{h}- \frac{e^h}{h}\right) -2x - h \right]$$

And now if you know how to develop the exponential, it should be straightforward.

What does it mean to "develop" the exponential? I apologize, I never really understood math terminology. Words like argument and stuff... always go beyond me

Thanks

 

[Amentia]

What I mean is to use this formula:

$$e^h = 1+h+h^2/2 + h^3/3! + ... = \sum_{n=0}^{\infty} h^n/n!$$

But you only need:

$$e^h = 1 + h +o(h^2)$$

because all the higher terms will vanish when you set h = 0.

 

[MrDickinson]

What I mean is to use this formula:

$$e^h = 1+h+h^2/2 + h^3/3! + ... = \sum_{n=0}^{\infty} h^n/n!$$

But you only need:

$$e^h = 1 + h +o(h^2)$$

because all the higher terms will vanish when you set h = 0.

Is this a series? I have not taken calculus 2... is there a way to simplify this limit without using a series? I am not really familiar with them...

 

[Amentia]

Yes, and it must be possible but right now I don't remember another way to simplify.

 

[MrDickinson]

Yes, and it must be possible but right now I don't remember another way to simplify.

Thank you.This problem is a practice problem to develop the quotient rule for differentiation (I have only taken calculus 1 and am waiting to take calculus 2). I decided that I would like to try to use the limit definition to find the derivative for this particular function (I usually pick complicated looking functions to really challenge my limited capacities). Thanks for all of the help though!