If f(x) has a derivative at x = a, then matho | vedclass

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

Question

(A) Given the limit: $\mathop {\lim }\limits_{x 	o a} \frac{xf(a) - af(x)}{x - a}$.To evaluate this,we add and subtract $af(a)$ in the numerator:$= \

Vedclass · Accepted Answer

$f(a) - a f'(a)$

Vedclass · Answer

(A) Given the limit: $\mathop {\lim }\limits_{x 	o a} \frac{xf(a) - af(x)}{x - a}$.To evaluate this,we add and subtract $af(a)$ in the numerator:$= \mathop {\lim }\limits_{x 	o a} \frac{xf(a) - af(a) + af(a) - af(x)}{x - a}$$= \mathop {\lim }\limits_{x 	o a} \left[ \frac{f(a)(x - a)}{x - a} - \frac{a(f(x) - f(a))}{x - a} ight]$$= \mathop {\lim }\limits_{x 	o a} f(a) - a \mathop {\lim }\limits_{x 	o a} \frac{f(x) - f(a)}{x - a}$Since $f(x)$ is differentiable at $x = a$,$\mathop {\lim }\limits_{x 	o a} \frac{f(x) - f(a)}{x - a} = f'(a)$.Therefore,the limit is $f(a) - af'(a)$.

If $f(x)$ has a derivative at $x = a,$ then $\mathop {\lim }\limits_{x \to a} \frac{xf(a) - af(x)}{x - a}$ is equal to

Similar Questions

Differentiate $\sin \left(\cos \left(x^{2}\right)\right)$ with respect to $x.$

Let $h(x)$ be differentiable for all $x$ and let $f(x) = (kx + e^x) h(x)$ where $k$ is some constant. If $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,then the value of $k$ is equal to

If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \dots$

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

Let $f$ be any function defined on $R$ and let it satisfy the condition $|f(x) - f(y)| \leq |(x - y)^2|$,for all $(x, y) \in R$. If $f(0) = 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module