Step-by-step explanation:
Letting the variable x approach 1 will give us [tex]\frac{0}{0}[/tex], an indeterminate result. In such a case, we can use L'Hopital's rule where
[tex]\displaystyle \lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f'(x)}{g'(x)}[/tex]
Let [tex]f(x) = \sqrt{17 - x} - 4[/tex] and
[tex]g(x) = \sqrt{ 2 - x} - 1[/tex]
We can see that
[tex]f'(x) = -\frac{1}{2}(17 - x)^{-\frac{1}{2}}[/tex]
and
[tex]g'(x) = -\frac{1}{2}(2 - x)^{-\frac{1}{2}}[/tex]
Applying L'Hopital's rule, we get
[tex]\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}}}[/tex]
[tex]\:\:\:\:\:\:\:\:= \displaystyle \lim_{x \to 1} \sqrt{\dfrac{2 - x}{17 - x}} =\sqrt{\dfrac{1}{16}} = \dfrac{1}{4}[/tex]
Therefore, as [tex]x \rightarrow 1,[/tex] the given expression approaches [tex]\frac{1}{4}[/tex].
