# Limit problem using l'hopital's 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?

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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 .

HallsofIvy
Homework Helper
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 ...

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.

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