# Math Help - L'Hospital's Rule: Two Problems?

1. ## L'Hospital's Rule: Two Problems?

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

2. The X-> represents as x goes to

3. Originally Posted by DocileRaptor
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,

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

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

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, $\sqrt{x^2+x}+x$. Then you would be able to use L'Hospital's rule.

4. 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?

$\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.

5. For the second one, I end up with...

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

Not sure how this is equal to 1/2

6. Originally Posted by DocileRaptor
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?

$\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,

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

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

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.

7. Originally Posted by DocileRaptor
For the second one, I end up with...

$
\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,

$\lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x }$

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.....

8. Originally Posted by Sudharaka
Dear DocileRaptor,

By L'Hospital's rule,

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

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

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

$\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

9. Originally Posted by DocileRaptor
Ah ha! I see now. This is the first day with L'Hospital's Rule.

So using L'Hospital's Rule twice consecutively on

$\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,

$\lim_{x\rightarrow{1}}\left(\frac{x\ln{x}-x+1}{(x-1)\ln{x}}\right)
=\lim_{x\rightarrow{1}}\left(\frac{\frac{1}{x}}{1+ \frac{1}{x}}\right)=\frac{1}{2}$

10. Originally Posted by Sudharaka
Dear DocileRaptor,

You are correct. The limit becomes,

$\lim_{x\rightarrow{\infty}}\frac{x}{\sqrt{x^2+x}+x }$

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.....

Can you show me what happens when you divide by X?

11. Originally Posted by DocileRaptor
Can you show me what happens when you divide by X?

12. The top divided by x is x/x = 1
the bottom is (sqrt(x^2 + x) / x ) + (x/x)

So it becomes 1/(sqrt(x^2 + x) / x ) +1

13. Originally Posted by DocileRaptor
The top divided by x is x/x = 1
the bottom is (sqrt(x^2 + x) / x ) + (x/x)

So it becomes 1/(sqrt(x^2 + x) / x ) +1
Dear DocileRaptor,

If you simplify more,

$\frac{1}{\frac{\sqrt{x^2+x}}{x}+1}=\frac{1}{\frac{ \sqrt{x^2+x}}{\sqrt{x^2}}+1}=\frac{1}{\sqrt{1+\fra c{1}{x}}+1}$

Hope you would be able to continue from here.

14. Doesn't the bottom become sqrt(2) + 1 ?

I feel like I'm missing something...

15. Originally Posted by DocileRaptor
Doesn't the bottom become sqrt(2) + 1 ?

I feel like I'm missing something...
Dear DocileRaptor,

$\lim_{x\rightarrow{\infty}}\frac{1}{\sqrt{1+\frac{ 1}{x}}+1}$

$=\frac{1}{\sqrt{1+0}+1}=\frac{1}{1+1}=\frac{1}{2}$

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