[tex]\displaystyle \lim_{x \to 4} \dfrac{x-4}{\sqrt{x} -2} = 4[/tex]
Step-by-step explanation:
[tex]\displaystyle \lim_{x \to 4} \dfrac{x-4}{\sqrt{x} -2}[/tex]
A straightforward substitution of [tex]x=4[/tex] will give us an indeterminate solution. So we use L'Hopital's rule which states that
[tex]\displaystyle \lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f'(x)}{g'(x)} [/tex]
where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively. We can see that
[tex]f'(x) = 1\:\:\:\: \text{and} \:\:\:\:g'(x)= \frac{1}{2} \frac{1}{\sqrt{x}}[/tex]
Therefore,
[tex]\displaystyle \lim_{x \to 4} \dfrac{x-4}{\sqrt{x} -2} = \lim_{x \to 4} 2\sqrt{x} = 4[/tex]
