# Limit without using L'Hopital's rule.

## Homework Statement

$$\lim_{x\rightarrow 1} {\frac{\sqrt{x}-1}{\sqrt{x}-1}}$$

I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?

Last edited:

Curious3141
Homework Helper

## Homework Statement

$$\lim_{x\rightarrow 1} {\frac{\sqrt{x}-1}{\sqrt{x}-1}}$$

I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?

One way is to let ##x = (y+1)^2##, after which the numerator can be simplified with the generalised binomial theorem.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

$$\lim_{x\rightarrow 1} {\frac{\sqrt{x}-1}{\sqrt{x}-1}}$$I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?
Do you know how to factor the difference of squares and the difference cubes ?

$(a-b)(a+b)=a^2-b^2$

$(c-d)(c^2+cd+d^2)=c^3-d^3$

Therefore, $\ \ (\sqrt{x\,}-1)(\sqrt{x\,}+1)=x-1$

and $\ \ (\sqrt{x}-1)(\sqrt{x^2}+\sqrt{x}+1)=x-1\ .$

These can be used to cancel x-1 in the numerator & denominator.

Thanks, both responses were very helpful.