Solving a limit by l'hopital's rule

[iwantcalculus]

So, according to answer sheet, the answer is 1...

The question is :

limit as x approaches infinity of : squareroot( x^2 + x ) - squareroot( x^2 - x)

I tried to put it in a limit calculator, but the steps shown are very complex and don't even involve l'hopital's rule...

I think the solution is by making e (natural number) to the power of natural logarithim of the function, but it's not working with me.. when I do that, I get the final answer e, but the final answer should be 1, as shown by answer sheet and limit calculator... please help...

 

[Orodruin]

You need to show us what you actually did, not just describe it in words. How else are we going to see where you go wrong? Also, please do not delete the homework template. It is provided for a reason.

 

[iwantcalculus]

You need to show us what you actually did, not just describe it in words. How else are we going to see where you go wrong? Also, please do not delete the homework template. It is provided for a reason.

I thought you were supposed to read the template and then remove it, but I did the stuff in the template... Now I hope mods don't remove it...

And this is my solution :

please help!

 

[Samy_A]

An error in your calculation: ##\log(a-b) \neq \frac{\log(a)}{\log(b)}##

(A correct equation is ##\log(\frac {a}{b})=\log(a)-\log(b)##.)

What happens when you multiply and divide ##(\sqrt{x²+x}-\sqrt{x²-x})## with ##(\sqrt{x²+x}+\sqrt{x²-x})##?

 

[Orodruin]

Please don't attach images like that, type it out in the forum! There are several reasons for this, one being that it is difficult to read, another that it is impossible to quote.

You seem to have used ##\ln(x-y) = \ln(x)/\ln(y)##. This is not true, what is true is ##\ln(x/y) = \ln(x) - \ln(y)##.

 

[iwantcalculus]

An error in your calculation: ##\log(a-b) \neq \frac{\log(a)}{\log(b)}##

(A correct equation is ##\log(\frac {a}{b})=\log(a)-\log(b)##.)

What happens when you multiply and divide ##(\sqrt{x²+x}-\sqrt{x²-x})## with ##(\sqrt{x²+x}+\sqrt{x²-x})##?

Please don't attach images like that, type it out in the forum! There are several reasons for this, one being that it is difficult to read, another that it is impossible to quote.

You seem to have used ##\ln(x-y) = \ln(x)/\ln(y)##. This is not true, what is true is ##\ln(x/y) = \ln(x) - \ln(y)##.

Sorry, so how do you solve these questions correctly if that property is not true? I am stuck there now...

 

[Orodruin]

Sorry, so how do you solve these questions correctly if that property is not true? I am stuck there now...

Samy_A also asked you a direct question in his post. Try to answer that.

 

[Samy_A]

Sorry, so how do you solve these questions correctly if that property is not true? I am stuck there now...

L'Hôpital's rule applies to a fraction. So you have to transform ##(\sqrt{x²+x}-\sqrt{x²-x})## into a fraction.
One way to do this is what I suggested in my previous post:

What happens when you multiply and divide ##(\sqrt{x²+x}-\sqrt{x²-x})## with ##(\sqrt{x²+x}+\sqrt{x²-x})##?

 

[iwantcalculus]

L'Hôpital's rule applies to a fraction. So you have to transform ##(\sqrt{x²+x}-\sqrt{x²-x})## into a fraction.
One way to do this is what I suggested in my previous post:

Thanks alot... problem solved! turns out there was no need for e and ln...\Edit: a new question came to my mind,

about the natural log, when we take natural log of both sides

like imagine : x+x = x^2 +x -2 or anything you can make up
if i take natural log of both sides

is it ln(x+x) = ln(x^2 +x -2)

or

lnx+lnx = lnx^2 + ln x - ln2

 

[Samy_A]

Thanks alot... problem solved! turns out there was no need for e and ln...\
about the natural log, when we take natural log of both sides

like imagine : x+x = x^2 +x -2 or anything you can make up
if i take natural log of both sides

is it ln(x+x) = ln(x^2 +x -2)

or

lnx+lnx = lnx^2 + ln x - ln2

The first one.
Some basic and useful properties of the logarithm (to any base) are:
##\log(ab)=\log(a)+\log(b)##
##\log(\frac {a}{b})=\log(a)-\log(b)##
##\log(a^b)=b\log(a)##