Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
[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]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.