If the function f defined as f(x) = frac{1}{x | vedclass

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

Question

(A) If the function is continuous at $x = 0$,then $\lim_{x 	o 0} f(x)$ must exist and $f(0) = \lim_{x 	o 0} f(x)$.Now,$\lim_{x 	o 0} f(x) = \lim_{x

Vedclass · Accepted Answer

$(3, 1)$

Vedclass · Answer

(A) If the function is continuous at $x = 0$,then $\lim_{x 	o 0} f(x)$ must exist and $f(0) = \lim_{x 	o 0} f(x)$.Now,$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} ight)$.$= \lim_{x 	o 0} \left( \frac{e^{2x} - 1 - (k - 1)x}{x(e^{2x} - 1)} ight)$.Using the expansion $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$,we get:$= \lim_{x 	o 0} \frac{(1 + 2x + 2x^2 + \dots) - 1 - kx + x}{x(2x + 2x^2 + \dots)}$.$= \lim_{x 	o 0} \frac{(3 - k)x + 2x^2 + \dots}{2x^2 + 2x^3 + \dots}$.For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $3 - k = 0$,which gives $k = 3$.Substituting $k = 3$,the limit becomes $\lim_{x 	o 0} \frac{2x^2 + \dots}{2x^2 + 2x^3 + \dots} = \frac{2}{2} = 1$.Therefore,$f(0) = 1$,and the ordered pair is $(3, 1)$.

If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

Similar Questions

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?

Which one of the following functions is discontinuous at $x=1$?

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when

Discuss the continuity of the function $f$ given by
$f(x) = \begin{cases} x, & \text{if } x \ge 0 \\ x^2, & \text{if } x < 0 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module