L'Hospital's Rule: Two Problems?

May 2010
10
0
I've been working on these two problems for some time and cannot get either of them to work. ( I don't know how to use Math HTML so bear with me )

Find the Limit

Problem 1: lim [ (x/x-1) - (1/lnx) ]
x->1

Problem 2: lim [ sqrt(x^2 + x) - x ]
x->inf


I've been working for quite a while. The solution to both problems is 1/2
 
Dec 2009
872
381
1111
I've been working on these two problems for some time and cannot get either of them to work. ( I don't know how to use Math HTML so bear with me )

Find the Limit

Problem 1: lim [ (x/x-1) - (1/lnx) ]
x->1

Problem 2: lim [ sqrt(x^2 + x) - x ]
x->inf


I've been working for quite a while. The solution to both problems is 1/2
Dear DocileRaptor,

For the first problem,

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x}{x-1}-\frac{1}{\ln{x}}\right)\)

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)\)

When x=1 the numerator and denominator becomes zero. Hence you can use the L'Hospitals rule. Hope you can continue from here.

For the second problem, multiply both the numerator and denominator by, \(\displaystyle \sqrt{x^2+x}+x\). Then you would be able to use L'Hospital's rule.

Hope this will help you.
 
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May 2010
10
0
Hello Sudharaka,

I suppose I should have posted how far I did get. I got to this step.
However, when I tried using L'Hospital's Rule, I end up with 0/2 which is incorrect. Can you show me a step process of solving this problem?

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

I'll try that for the second one, multiplying by the reciprocal makes sense.
 
May 2010
10
0
For the second one, I end up with...

\(\displaystyle
\lim_{x\rightarrow{\infty}}\left(\frac{x}{\sqrt{x^2+x}+x}\right) \)



Not sure how this is equal to 1/2
 
Dec 2009
872
381
1111
Hello Sudharaka,

I suppose I should have posted how far I did get. I got to this step.
However, when I tried using L'Hospital's Rule, I end up with 0/2 which is incorrect. Can you show me a step process of solving this problem?

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

I'll try that for the second one, multiplying by the reciprocal makes sense.
Dear DocileRaptor,

By L'Hospital's rule,

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

\(\displaystyle =\lim_{x\rightarrow{1}}\left(\frac{1+\ln{x}-1}{\frac{x-1}{x}+\ln{x}}\right)\)

When x=1, the numerator and denominator both becomes to zero. Hence you have to use the L'Hospital's rule again. Hope you can continue from here.
 
Dec 2009
872
381
1111
For the second one, I end up with...

\(\displaystyle
\lim_{x\rightarrow{\infty}}\left(\frac{x}{\sqrt{x^2+x}+x}\right) \)



Not sure how this is equal to 1/2
Dear DocileRaptor,

You are correct. The limit becomes,

\(\displaystyle \lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x}\)

Now you can use the L'Hospital's rule(since when x goes to infinity both the numerator and denominator goes to infinity).

But there's an easy method. Divide both the numerator and denominator by x.....

Hope this will help you.
 
May 2010
10
0
Dear DocileRaptor,

By L'Hospital's rule,

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

\(\displaystyle =\lim_{x\rightarrow{1}}\left(\frac{1+\ln{x}-1}{\frac{x-1}{x}+\ln{x}}\right)\)

When x=1, the numerator and denominator both becomes to zero. Hence you have to use the L'Hospital's rule again. Hope you can continue from here.
Ah ha! I see now. This is the first day with L'Hospital's Rule.

So using L'Hospital's Rule twice consecutively on

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

I end up with

(1/x) / (1+ (1/x) ) as X approaches 1 which is equal to 1/2 Can you put that equation into the code so I can see how the code looks? I am learning how to use latex as we speak :)
 
Dec 2009
872
381
1111
Ah ha! I see now. This is the first day with L'Hospital's Rule.

So using L'Hospital's Rule twice consecutively on

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
\)

I end up with

(1/x) / (1+ (1/x) ) as X approaches 1 which is equal to 1/2 Can you put that equation into the code so I can see how the code looks? I am learning how to use latex as we speak :)
Dear DocileRaptor,

\(\displaystyle \lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
=\lim_{x\rightarrow{1}}\left(\frac{\frac{1}{x}}{1+\frac{1}{x}}\right)=\frac{1}{2}\)
 
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Reactions: DocileRaptor
May 2010
10
0
Dear DocileRaptor,

You are correct. The limit becomes,

\(\displaystyle \lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x}\)

Now you can use the L'Hospital's rule(since when x goes to infinity both the numerator and denominator goes to infinity).

But there's an easy method. Divide both the numerator and denominator by x.....

Hope this will help you.
Can you show me what happens when you divide by X?