# How can I find the limit of (sqrt(16-x)-4)/(x) as it approaches to 0?

Jul 18, 2016

${\lim}_{x \to 0} \frac{\sqrt{16 - x} - 4}{x} = - \frac{1}{8}$

#### Explanation:

For this problem, we can make use of some properties of limits, which will come in handy once we try to evaluate the limit.

First, however, we notice that direct substitution yields the indeterminate form of $\frac{0}{0}$. We now have two options:

1. Do it algebraically
2. Use L'Hopital's rule

We will start with the first option, which is doing it algebraically.

The "Algebraic" Way:

Before we do anything with the limit, we can make use of what's also known as the conjugate. The conjugate is basically just the same expression with subtraction switched to addition or vice versa. In our case, we would like to get rid of the square root in the numerator.

Thus, we can write

${\lim}_{x \to 0} \frac{\sqrt{16 - x} - 4}{x} = {\lim}_{x \to 0} \frac{\sqrt{16 - x} - 4}{x} \cdot \frac{\sqrt{16 - x} + 4}{\sqrt{16 - x} + 4}$

However, the product of conjugates is always the square of the first term minus the square of the second term, so we write

 lim_(x->0) (sqrt(16-x)-4)/(x) * (sqrt(16-x)+4)/ (sqrt(16-x)+4) = lim_(x->0) (16-x-4^2)/(x(sqrt(16-x)+4)

= lim_(x->0) (16-x-16)/(x(sqrt(16-x)+4)) = lim_(x->0) (-cancel(x))/(cancel(x)(sqrt(16-x)+4)

$= {\lim}_{x \to 0} - \frac{1}{\sqrt{16 - x} + 4} = - \frac{1}{\sqrt{16 - 0} + 4} = - \frac{1}{8}$

Using L'Hospital's rule:

The basic idea of L'Hospital's rule is that (there needs to be!) an indeterminate form such as $\frac{0}{0}$ in order for it to work. In our case, direct substitution of $x = 0$ yields this form, thus we can start applying L'Hospital's.

L'Hospital's rule tells us that if such an indeterminate form exists, we can take the derivative of the numerator and denominator - which in our case gives us the following:

${\lim}_{x \to 0} \frac{\sqrt{16 - x} - 4}{x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(\sqrt{16 - x} - 4\right)}{\frac{d}{\mathrm{dx}} \left(x\right)}$

$= {\lim}_{x \to 0} \frac{\frac{1}{2} {\left(16 - x\right)}^{- \frac{1}{2}} \left(- 1\right)}{1}$

$= {\lim}_{x \to 0} - \frac{1}{2 \sqrt{16 - x}}$

$= - \frac{1}{2 \sqrt{16 - 0}} = - \frac{1}{8}$

We can also check if this answer makes sense by graphing our original function.

graph{(sqrt(16-x)-4)/(x) [-0.843, 0.843, -0.4215, 0.4214]}