Given that f'(2) = 6 and f'(1) = 4,then matho | vedclass

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

Question

(D) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $h$:$\mathop {\lim }\limits_{h 	o 0} \frac{{f(2h + 2 + {h

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $h$:$\mathop {\lim }\limits_{h 	o 0} \frac{{f(2h + 2 + {h^2}) - f(2)}}{{f(h - {h^2} + 1) - f(1)}} = \mathop {\lim }\limits_{h 	o 0} \frac{{f'(2h + 2 + {h^2}) \cdot (2 + 2h)}}{{f'(h - {h^2} + 1) \cdot (1 - 2h)}}$As $h 	o 0$,the expression becomes:$\frac{{f'(2) \cdot 2}}{{f'(1) \cdot 1}} = \frac{{6 	imes 2}}{{4 	imes 1}} = \frac{{12}}{{4}} = 3$.

Given that $f'(2) = 6$ and $f'(1) = 4$,then $\mathop {\lim }\limits_{h \to 0} \frac{{f(2h + 2 + {h^2}) - f(2)}}{{f(h - {h^2} + 1) - f(1)}} = $

Similar Questions

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

Evaluate the limit: $\lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = $

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module