Question

Trying to simplify an equation, but it's apparently wrong.

Answer 1

Step-by-step explanation:

Letting the variable x approach 1 will give us $\frac{0}{0}$, an indeterminate result. In such a case, we can use L'Hopital's rule where

$\displaystyle \lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f'(x)}{g'(x)}$

Let $f(x) = \sqrt{17 - x} - 4$ and

$g(x) = \sqrt{ 2 - x} - 1$

We can see that

$f'(x) = -\frac{1}{2}(17 - x)^{-\frac{1}{2}}$

and

$g'(x) = -\frac{1}{2}(2 - x)^{-\frac{1}{2}}$

Applying L'Hopital's rule, we get

$\displaystyle \lim_{x \to 1} \dfrac{\sqrt{17 - x} - 4}{\sqrt{ 2 - x} - 1} = \lim_{x \to 1} \dfrac{-\frac{1}{2}(17 - x)^{-\frac{1}{2}}}{-\frac{1}{2}(2 - x)^{-\frac{1}{2}}}$

$\:\:\:\:\:\:\:\:= \displaystyle \lim_{x \to 1} \sqrt{\dfrac{2 - x}{17 - x}} =\sqrt{\dfrac{1}{16}} = \dfrac{1}{4}$

Therefore, as $x \rightarrow 1,$ the given expression approaches $\frac{1}{4}$.