# L'Hospital's Rule: Two Problems?

#### 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

#### DocileRaptor

The X-> represents as x goes to

#### Sudharaka

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.

DocileRaptor

#### 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?

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

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

#### DocileRaptor

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

#### Sudharaka

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.

#### Sudharaka

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.....

#### DocileRaptor

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

#### Sudharaka

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,

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

DocileRaptor

#### DocileRaptor

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.....