Chapter 1: Problem 41
$\lim _{n \rightarrow \infty}\left(\frac{\sqrt[n]{p}+\sqrt[n]{q}}{2}\right)^{n}\(, p, \)q>0$ is equal to (A) 1 (B) \(\sqrt{\mathrm{pq}}\) (C) pq (D) \(\frac{\mathrm{pq}}{2}\)
Chapter 1: Problem 41
$\lim _{n \rightarrow \infty}\left(\frac{\sqrt[n]{p}+\sqrt[n]{q}}{2}\right)^{n}\(, p, \)q>0$ is equal to (A) 1 (B) \(\sqrt{\mathrm{pq}}\) (C) pq (D) \(\frac{\mathrm{pq}}{2}\)
All the tools & learning materials you need for study success - in one app.
Get started for free$\lim _{x \rightarrow 0} \frac{\ell n\left(1+x+x^{2}\right)+\ell n\left(1-x+x^{2}\right)}{\sec x-\cos x}$ is equal to (A) 1 (B) \(-1\) (C) 0 (D) \(\infty\)
Give a real valued function \(\mathrm{f}\) such that $f(x)=\left\\{\begin{array}{cll}\frac{\tan ^{2} x}{\left(x^{2}-[x]\right)^{2}} & \text { for } & x>0 \\ 1 & \text { for } & x=0 \\ \sqrt{\\{x\\} \cot \\{x\\}} & \text { for } & x<0\end{array}\right.$ where, [.] is the integral part and \\{\\} is the fractional part of \(\mathrm{x}\), then (A) \(\lim _{x \rightarrow 0} f(x)=1\) (B) \(\lim _{x \rightarrow 0^{-}} f(x)=\cot 1\) (C) \(\cot ^{-1}\left(\lim _{x \rightarrow 0^{-}} f(x)\right)^{2}=1\) (D) none of these
Assertion \((\mathbf{A}):\) Let \(\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}\) be a twice continuously differentiable function such that $\left|f^{\prime}(x)+2 x f^{\prime}(x)+\left(x^{2}+1\right) f(x)\right| \leq 1\( for all \)x$ Then \(\lim _{x \rightarrow \infty} f(x)=0\). Reason (R) : Applying L'Hospital's rule twice on the function $\frac{f(x) e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}}\( we get \)\lim _{x \rightarrow \infty} f(x)=0$.
$\lim _{x \rightarrow 1} \frac{(\ell n(1+x)-\ell \operatorname{n} 2)\left(3.4^{x-1}-3 x\right)}{\left[(7+x)^{1 / 3}-(1+3 x)^{1 / 2}\right] \cdot \sin (x-1)}$ equals (A) \(\frac{9}{4}\) en \(\frac{4}{\mathrm{e}}\) (B) \(\frac{9}{4}\) en \(\frac{\mathrm{e}}{4}\) (C) \(\frac{4}{9} \ell \mathrm{n} \frac{\mathrm{e}}{4}\) (D) None of these
$\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{\mathrm{n}}{\mathrm{n}^{2}-2}+\frac{4^{\mathrm{n}}(-1)^{\mathrm{n}}}{2^{\mathrm{n}}-1}\right)^{-1}$ is equal to (A) 2 (B) (C) 0 (D) None of these
What do you think about this solution?
We value your feedback to improve our textbook solutions.