# How do you find the limit lim_(h->0)(sqrt(1+h)-1)/h ?

May 19, 2018

$\setminus \frac{1}{2}$

#### Explanation:

The limit presents an undefined form $\frac{0}{0}$. In this case, you may use de l'hospital theorem, that states

$\lim \setminus \frac{f \left(x\right)}{g \left(x\right)} = \setminus \lim \setminus \frac{f ' \left(x\right)}{g ' \left(x\right)}$

The derivative of the numerator is

$\setminus \frac{1}{2 \sqrt{1 + h}}$

While the derivative of the denominator is simply $1$.

So,

$\setminus {\lim}_{x \setminus \to 0} \setminus \frac{f ' \left(x\right)}{g ' \left(x\right)} = \setminus {\lim}_{x \setminus \to 0} \setminus \frac{\setminus \frac{1}{2 \sqrt{1 + h}}}{1} = \setminus {\lim}_{x \setminus \to 0} \setminus \frac{1}{2 \sqrt{1 + h}}$

And thus simply

$\setminus \frac{1}{2 \sqrt{1}} = \setminus \frac{1}{2}$

May 19, 2018

$= \frac{1}{2}$

#### Explanation:

If you are unaware of l'hopitals rule...

Use:

(1+x)^n = 1 + nx + (n(n-1))/(2!) x^2 + ...

$\implies {\left(1 + h\right)}^{\frac{1}{2}} = 1 + \frac{1}{2} h - \frac{1}{8} {h}^{2} + \ldots$

$\implies {\lim}_{h \to 0} \frac{\left(1 + \frac{1}{2} h - \frac{1}{8} {h}^{2} + \ldots\right) - 1}{h}$

$\implies {\lim}_{h \to 0} \frac{\frac{1}{2} h - \frac{1}{8} {h}^{2} + \ldots}{h}$

$\implies {\lim}_{h \to 0} \left(\frac{1}{2} - \frac{1}{8} h + \ldots\right)$

$= \frac{1}{2}$