If f(x) = begin{cases} frac{a}{2}(x - |x|), & | vedclass

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

Question

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = f(0) = \lim_{x 	o 0^{+}} f(x)$.First,consider the left-hand limit:

Vedclass · Accepted Answer

$a$ is any real value and $b$ is any real value

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = f(0) = \lim_{x 	o 0^{+}} f(x)$.First,consider the left-hand limit: $\lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0^{-}} \frac{a}{2}(x - |x|)$.Since $x This holds true for any real value of $a$.Next,consider the right-hand limit: $\lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0^{+}} bx^2 \sin \left( \frac{1}{x} ight)$.We know that $-1 \leq \sin \left( \frac{1}{x} ight) \leq 1$.Multiplying by $bx^2$,we get $-|bx^2| \leq bx^2 \sin \left( \frac{1}{x} ight) \leq |bx^2|$.By the Squeeze Theorem,as $x 	o 0$,$bx^2 \sin \left( \frac{1}{x} ight) 	o 0$.This holds true for any real value of $b$.Thus,$f(x)$ is continuous at $x = 0$ for any real values of $a$ and $b$.

If $f(x) = \begin{cases} \frac{a}{2}(x - |x|), & \text{for } x < 0 \\ 0, & \text{for } x = 0 \\ bx^2 \sin \left( \frac{1}{x} \right), & \text{for } x > 0 \end{cases}$ is continuous at $x = 0$,then

Similar Questions

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

Let $f(x) = \min \{x, x^2\}$ for every real number $x$. Then:

If $f(x) = \frac{x+x^2+x^3+\ldots+x^{n}-n}{x-1}$ for $x \neq 1$ is continuous at $x=1$,then $f(1) =$

If $f(x) = \begin{cases} \frac{1-\cos Kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then the value of $K$ is

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module