If f(x) = begin{cases} frac{(e^{3x}-1) sin x^ | vedclass

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

Question

(A) To check continuity at $x=0$,we evaluate $\lim_{x ightarrow 0} f(x)$.Given $f(x) = \frac{(e^{3x}-1) \sin x^{\circ}}{x^2}$ for $x 
eq 0$.We know

Vedclass · Accepted Answer

$f$ is continuous at $x=0$

Vedclass · Answer

(A) To check continuity at $x=0$,we evaluate $\lim_{x ightarrow 0} f(x)$.Given $f(x) = \frac{(e^{3x}-1) \sin x^{\circ}}{x^2}$ for $x 
eq 0$.We know that $x^{\circ} = \frac{\pi x}{180}$ radians.So,$\lim_{x ightarrow 0} f(x) = \lim_{x ightarrow 0} \frac{(e^{3x}-1) \sin(\frac{\pi x}{180})}{x^2}$.Multiplying and dividing by $3x$ and $\frac{\pi}{180}$,we get:$\lim_{x ightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 ight) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} ight) \cdot \frac{1}{x} \cdot x^2$ is not correct,let us simplify:$\lim_{x ightarrow 0} \frac{e^{3x}-1}{x} \cdot \frac{\sin(\frac{\pi x}{180})}{x} = \lim_{x ightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 ight) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} ight) = (1 \cdot 3) \cdot (1 \cdot \frac{\pi}{180}) = \frac{3\pi}{180} = \frac{\pi}{60}$.Since $\lim_{x ightarrow 0} f(x) = f(0) = \frac{\pi}{60}$,the function $f(x)$ is continuous at $x=0$.

If $f(x) = \begin{cases} \frac{(e^{3x}-1) \sin x^{\circ}}{x^2} & x \neq 0 \\ \frac{\pi}{60} & x = 0 \end{cases}$,then:

Similar Questions

If $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

If $f(x) = \begin{cases} \frac{x-2}{|x-2|}+a & , x<2 \\ a+b & , x=2 \\ \frac{x-2}{|x-2|}+b & , x>2 \end{cases}$ is continuous at $x=2$, then $a+b=$

Let $k$ be a non-zero real number. If $f(x) = \begin{cases} \frac{(e^x - 1)^2}{\sin (x/k) \log (1 + x/4)}, & x \neq 0 \\ 12, & x = 0 \end{cases}$ is a continuous function,then the value of $k$ is

If the function $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x \ne 0$ and $f(x) = k$ for $x = 0$ is a continuous function at $x = 0$,then the value of $k$ is:

Consider the function $f(x) = \begin{cases} \frac{x+5}{x-2}, & \text{if } x \neq 2 \\ 1, & \text{if } x=2 \end{cases}$. Then,$f(f(x))$ is discontinuous

Vedclass Test Series

Exam Paper Generator

Online Exam Module