# How do you find lim (sqrt(x+1)-1)/(sqrt(x+2)-1) as x->0 using l'Hospital's Rule or otherwise?

##### 2 Answers
Dec 14, 2016

L'Hôpital's rule should not be used. Determine the limit by evaluating at $x = 0$.

#### Explanation:

Because the expression evaluated at the limit does not create an indeterminate form, then L'Hôpital's rule should not be used.

The limit can be determined by evaluation at $x = 0$:

${\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 2} - 1} = \frac{\sqrt{0 + 1} - 1}{\sqrt{0 + 2} - 1} = \frac{1 - 1}{\sqrt{2} - 1} = 0$

Dec 14, 2016

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 2} - 1} = 0$

#### Explanation:

L'Hôpital's rule can only be applied to a limit of an indeterminate form $\frac{0}{0}$ (or equivalently $\frac{\infty}{\infty}$).

For the given limit

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 2} - 1}$

we note that the numerator$= 0$ but denominator$= 1$ so L'Hôpital's RuleCANNOT be used.

If we examine the graph of the function we can see it "looks like" the value of the limit is $0$
graph{y= (sqrt(x+1)-1)/(sqrt(x+2)-1) [-10, 10, -5, 5]}

In fact we can easily show this as the denominator is non-zero and so

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 2} - 1} = \frac{0}{1} = 0$

NB If you erroneously applied L'Hôpital's you would incorrectly get:

${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{\sqrt{x + 2} - 1} = {\lim}_{x \rightarrow 0} \frac{\frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}}}{\frac{1}{2} {\left(x + 2\right)}^{- \frac{1}{2}}}$

$\text{ } = {\lim}_{x \rightarrow 0} {\left(x + 1\right)}^{- \frac{1}{2}} / {\left(x + 2\right)}^{- \frac{1}{2}}$

$\text{ } = {\lim}_{x \rightarrow 0} \frac{\frac{1}{\sqrt{x + 1}}}{\frac{1}{\sqrt{x + 2}}}$

$\text{ } = {\lim}_{x \rightarrow 0} \frac{\sqrt{x + 2}}{\sqrt{x + 1}}$
$\text{ } = \frac{2}{1}$

$\text{ } = 2$

Which is complete nonsense.