To evaluate the limit
[tex]\[ \lim_{x \to 25} \frac{x - 25}{\sqrt{x} - 5}, \][/tex]
we note that both the numerator and the denominator approach 0 as [tex]\( x \)[/tex] approaches 25. This results in an indeterminate form [tex]\(\frac{0}{0}\)[/tex], making l'Hôpital's Rule applicable. l'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] is in indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, \][/tex]
provided this latter limit exists.
1. Differentiate the numerator and the denominator:
- The numerator [tex]\( f(x) = x - 25 \)[/tex] has the derivative [tex]\( f'(x) = 1 \)[/tex].
- The denominator [tex]\( g(x) = \sqrt{x} - 5 \)[/tex] can be differentiated using the chain rule. The derivative of [tex]\(\sqrt{x}\)[/tex] is [tex]\(\frac{1}{2\sqrt{x}}\)[/tex], thus
[tex]\[ g'(x) = \frac{1}{2\sqrt{x}}. \][/tex]
2. Apply l'Hôpital's Rule:
Substituting the derivatives we get:
[tex]\[ \lim_{x \to 25} \frac{x - 25}{\sqrt{x} - 5} = \lim_{x \to 25} \frac{1}{\frac{1}{2\sqrt{x}}}. \][/tex]
3. Simplify the expression:
[tex]\[ \frac{1}{\frac{1}{2\sqrt{x}}} = 2\sqrt{x}. \][/tex]
4. Evaluate the limit at [tex]\( x = 25 \)[/tex]:
[tex]\[ \lim_{x \to 25} 2\sqrt{x} = 2\sqrt{25} = 2 \times 5 = 10. \][/tex]
Thus, the limit is:
[tex]\[ \boxed{10}. \][/tex]
