Let f(x) = mathop {lim }limits_{n to infty } | vedclass

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

Question

(D) We analyze the limit $f(x) = \mathop {\lim }\limits_{n 	o \infty } \frac{{{{\left( {4\sin^2 x} ight)}^{n}}}}{{{3^n} - {{\left( {4\cos^2 x} ig

Vedclass · Accepted Answer

All the above statements are correct.

Vedclass · Answer

(D) We analyze the limit $f(x) = \mathop {\lim }\limits_{n 	o \infty } \frac{{{{\left( {4\sin^2 x} ight)}^{n}}}}{{{3^n} - {{\left( {4\cos^2 x} ight)}^{n}}}}$.Case $1$: If $|4\sin^2 x| This occurs when $\sin^2 x Case $2$: If $|4\sin^2 x| > 3$ and $|4\cos^2 x| This occurs when $\sin^2 x > \frac{3}{4}$,i.e.,$x \in (m\pi + \frac{\pi}{3}, m\pi + \frac{2\pi}{3})$.Case $3$: If $|4\cos^2 x| > 3$ and $|4\sin^2 x| At $x = m\pi \pm \frac{\pi}{6}$,the denominator approaches $0$ or the terms become undefined,leading to discontinuity. Thus,$f(x)$ is discontinuous at these points. Checking $f(0) = 0$ and $f(\pi/3) = 1$ confirms all statements are correct.

Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

Similar Questions

Let $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is:

Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is a continuous function.

If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

The function $f(x)=\sqrt{\frac{3 x^2-5 x-2}{2 x^2-7 x+5}}$ has discontinuous points at $x=$

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin x - \sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $f(0) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module