The value of f(0) so that the function f(x) = | vedclass

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

Question

(B) For $f(x)$ to be continuous at $x = 0$, $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(B) For $f(x)$ to be continuous at $x = 0$, $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}} = \lim_{x 	o 0} \frac{4(1 - \frac{8x}{256})^{\frac{1}{4}} - 4}{16 - 4 \cdot 4(1 + \frac{3x}{64})^{\frac{1}{3}}} = \lim_{x 	o 0} \frac{4(1 - \frac{x}{32})^{\frac{1}{4}} - 4}{16 - 16(1 + \frac{3x}{64})^{\frac{1}{3}}}$.Using the binomial approximation $(1 + u)^n \approx 1 + nu$ for small $u$:Numerator: $4(1 + \frac{1}{4} \cdot (-\frac{x}{32})) - 4 = 4 - \frac{x}{32} - 4 = -\frac{x}{32}$.Denominator: $16 - 16(1 + \frac{1}{3} \cdot \frac{3x}{64}) = 16 - 16 - \frac{16x}{64} = -\frac{x}{4}$.$\lim_{x 	o 0} \frac{-\frac{x}{32}}{-\frac{x}{4}} = \frac{4}{32} = \frac{1}{8}$.

The value of $f(0)$ so that the function $f(x) = \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^n} - 1}}{x} = $

$\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)=$

Evaluate the limit: $\lim _{x \rightarrow \infty}\left\{x-\sqrt[n]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}$, where $a_1, a_2, \ldots, a_n$ are positive rational numbers.

$\mathop {\lim }\limits_{n \to \infty } \sin (\pi \sqrt {{n^2} + 1} ) = $

The value of $\mathop {\lim }\limits_{x \to \infty } {x^{\frac{1}{3}}}\left( {{{\left( {x + 1} \right)}^{\frac{2}{3}}} - {{\left( {x - 1} \right)}^{\frac{2}{3}}}} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module