Question

How can I find the limit of $\dfrac{{\sqrt {16 - x} - 4}}{x}$ as it approaches 0?

Answer 1

L Hospital rule is a general method of evaluating indeterminate forms such as $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }$. To find the limit of the given expression we need to apply L'Hospital's rule, in which this rule uses the derivatives to evaluate the limits which involve the indeterminate forms and then simplify the expression and apply the given limit i.e.,$x \to 0$.

Complete step by step solution:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {16 - x} - 4}}{x}$
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 $\dfrac{0}{0}$.
Hence, we will do it using L'Hospital's rule:
The basic idea of L'Hospital's rule is that an indeterminate form such as $\dfrac{0}{0}$ 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:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {16 - x} - 4}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {\sqrt {16 - x} - 4} \right)}}{{\dfrac{d}{{dx}}\left( x \right)}}$
Now, find the derivative of the numerator and denominator terms and we know that $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{2}{x^{ - \dfrac{1}{2}}}$ and $\dfrac{d}{{dx}}\left( x \right) = 1$, hence applying this we get:
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{2}{{\left( {16 - x} \right)}^{ - \dfrac{1}{2}}}\left( { - 1} \right)}}{1}$
Evaluating the terms, we get:
$= \mathop {\lim }\limits_{x \to 0} - \dfrac{1}{{2\sqrt {16 - x} }}$
Applying the value of the given limits i.e., $x \to 0$, we get:
$= - \dfrac{1}{{2\sqrt {16 - 0} }}$
Evaluating the terms, we get:
$= - \dfrac{1}{{2\left( 4 \right)}}$
$= - \dfrac{1}{8}$

Hence, we get:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {16 - x} - 4}}{x} = - \dfrac{1}{8}$

Note: In L Hospital rule the limit of the quotient of function is equivalent to the limit of the quotient of their derivatives, given that the provided conditions are satisfied. If at a given point two functions have an infinite limit or zero as a limit and are both differentiable of this point then the limit of the quotient of the functions is equal to the limit of the quotient of their derivatives provided that this limit exists.