Chapter 1: Problem 1
The value of $\lim _{\mathrm{x} \rightarrow 0}\left[\frac{|\sin \mathrm{x}|}{|\mathrm{x}|}\right]$, (where [.] denotes greatest integer function) is (A) 0 (B) does not exists (C) \(-1\) (D) 1
Chapter 1: Problem 1
The value of $\lim _{\mathrm{x} \rightarrow 0}\left[\frac{|\sin \mathrm{x}|}{|\mathrm{x}|}\right]$, (where [.] denotes greatest integer function) is (A) 0 (B) does not exists (C) \(-1\) (D) 1
All the tools & learning materials you need for study success - in one app.
Get started for freeWhich of the following functions has two horizontal asymptotes (A) \(\mathrm{y}=\frac{|\mathrm{x}|}{\mathrm{x}+1}\) (B) \(\mathrm{y}=\frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}}\) (C) \(y=\frac{\sin x}{x^{2}+1}\) (D) \(y=\cot ^{-1}(2 x+1)\)
Column - I (A) \(\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{e^{x^{3}}-e^{x}+x}\) equals (B) If the value of $\lim _{\mathrm{x} \rightarrow 0^{-}}\left(\frac{(3 / \mathrm{x})+1}{(3 / \mathrm{x})-1}\right)^{1 / \mathrm{x}}$ can be expressed in the form of \(\mathrm{e}^{\mathrm{iq}}\), where \(\mathrm{p}\) and \(\mathrm{q}\) are relative prime then \((\mathrm{p}+\mathrm{q})\) is equal to (C) \(\lim _{x \rightarrow 0} \frac{\tan ^{3} x-\tan x^{3}}{x^{5}}\) equals (D) $\lim _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^{2}+2 \sin x+1}-\sqrt{\sin ^{2} x-x+1}}$ Column-II (P) 1 (Q) 2 (R) 4 (S) 5
Which of the following functions has a vertical asymptote at \(\mathrm{x}=-1\). (A) \(y=\frac{\left|x^{2}-1\right|}{x+1}\) (B) \(y=\frac{x^{2}-6 x-7}{x+1}\) (C) \(y=\frac{x^{2}+1}{x+1}\) (D) \(y=\frac{\sin (x+2)}{x+1}\)
If $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right)$ $=\mathrm{K} \cdot \lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})\( then the value of \)\mathrm{K}$ (A) (B) (C) \(2 \mathrm{a}\) (D) None of these
$\lim _{x \rightarrow 0} \lim _{x \rightarrow 0} \frac{2(\tan x-\sin x)-x^{3}}{x^{5}}$ is equal to (A) \(1 / 4\) (B) \(1 / 2\) (C) \(1 / 3\) (D) None of these
What do you think about this solution?
We value your feedback to improve our textbook solutions.