# Limit problem using l'hopital's rule

• mtayab1994
In summary, the problem was that you needed to do the derivative of the top over the derivative of the bottom right, but in this case you can just use the product rule.

## Homework Statement

$$\lim_{x\rightarrow2}\frac{x\sqrt{x-1}-2}{x-2}$$

## Homework Equations

calculate the limit

## The Attempt at a Solution

When I apply l'hopital's i have to do derivative of the top over derivative of the bottom right?

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mtayab1994 said:

## Homework Statement

$$\lim_{x\rightarrow2}\frac{x\sqrt{x-1}-2}{x-2}$$

## Homework Equations

calculate the limit

## The Attempt at a Solution

When I apply l'hopital's i have to do derivative of the top over derivative of the bottom right?
Yes .

You could use L'Hopital but I would consider that "overkill". Instead, rationalize the numerator and factor:

$$\frac{x\sqrt{x- 1}- 2}{x- 2}\frac{x\sqrt{x- 1}+ 2}{x\sqrt{x- 1}+ 2}= \frac{x^2(x- 1)- 4}{(x- 2)(x\sqrt{x- 1}+ 2)}= \frac{x^3- x^2- 4}{(x- 1)(x\sqrt{x- 1}+ 2)}$$
$$\frac{(x- 2)(x^2+ x+ 2)}{(x- 2)(x\sqrt{x- 1}- 2)}= \frac{x^2+ x+ 2}{x\sqrt{x- 1}+ 2}$$

Now, let x= 2.

(We knew it would factor that way precisely because x= 2 made both numerator and denominator equal to 0.)

Another alternative:

$$Let \ x \ =\ y^2 + 1.$$

$$As \ \ x\to 2, \ \ y \to 1.$$

$$\lim_{y \to 1} \dfrac{(y^2 + 1)\sqrt{y^2} - 2}{(y^2 + 1) - 2} \ =$$

$$\lim_{y \to 1}\dfrac{|y|(y^2 + 1) - 2}{y^2 - 1} \ = \$$

$$\lim_{y \to 1}\dfrac{y(y^2 + 1) - 2}{y^2 - 1} \ ** \ = \$$

$$\lim_{y \to 1}\dfrac{y^3 + y - 2}{y^2 - 1} \ = \$$

$$\lim_{y \to 1}\dfrac{(y - 1)(y^2 + y + 2)}{(y - 1)(y + 1)} \ = \$$

$$\lim_{y \to 1}\dfrac{y^2 + y + 2}{y + 1}$$

Continue . . .

$$** \ Note: \ \ \ As \ \ y \ \ is \ \ positive \ \ near \ \ 1, \ \ |y| = y.$$

I used L'Hopital's rule, and I think it's actually simpler in this case. The derivative of the denominator is 1, the derivative of the constant 2 is 0 so ...

zhandele said:
I used L'Hopital's rule,
and I think it's actually simpler in this case.

The derivative of the denominator is 1, $$\ \ \ \ Yes, \ \ so \ \ evidently \ \ you're \ \ done \ \ discussing \ \ the \ \ denominator.$$

the derivative of the constant 2 is 0 $$\ \ \ \ No, \ \ if \ \ you \ \ are \ \ now \ \ discussing \ \ the \ \ numerator,$$$$you \ \ are \ \ to \ \ be \ \ taking \ \ the \ \ derivative \ \ of \ \ \ x\sqrt{x - 1} - 2.$$
$$And \ \ that \ \ derivative\ \ is \ \ not \ \ equal \ \ to \ \ zero.$$

so ...

...

I meant that the derivative of the constant -2 (which appears in the numerator) is 0, so we just have to use the product rule to take the derivative of x*sqrt(x-1). The result,

√(x-1) + (x/2)*((√(x-1))^(-1))

... sorry I don't have practice entering these equations on this system ... but now you can just sub in x = 2 and the limit falls right out.

√(2-1)+(2/2)*((√(2-1)^(-1))

√1 + (1)*(1/√1)= 2

what could be simpler?

I hope I put in the right number of parens.

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Yes I solved it guys thanks for the help. By the way the problem stated on solving it using l'hopital's rule