If f(9) = 9 and f'(9) = 4,then mathop {lim }l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the limit $\mathop {\lim }\limits_{x 	o 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.Since $f(9) = 9$,the expression takes the indeterminat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the limit $\mathop {\lim }\limits_{x 	o 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.Since $f(9) = 9$,the expression takes the indeterminate form $\frac{0}{0}$ as $x 	o 9$.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 9} \frac{\frac{d}{dx}(\sqrt{f(x)} - 3)}{\frac{d}{dx}(\sqrt{x} - 3)} = \mathop {\lim }\limits_{x 	o 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x 	o 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$.Substituting the values $f(9) = 9$ and $f'(9) = 4$:$= \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

If $f(9) = 9$ and $f'(9) = 4$,then $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}} = $

Similar Questions

The value of $\mathop {\lim }\limits_{x \to {0^ + }} {x^m}{(\log x)^n}$,where $m, n \in N$,is

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha} = $

If $f: R \rightarrow R$ is such that $f(3)=16$ and $f^{\prime}(3)=4$,then find the value of $\lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$.

$\lim \limits_{x}$ ${\rightarrow a} \frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}} \text{ for } a \neq 0 \text{ is equal to}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module