Question
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:
-
$2$
-
$0$
-
$1$
-
$-1$
$\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:
$2$
$0$
$1$
$-1$
Solution:
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$
$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$
Taking limit, we get
$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]=\frac{1}{2}\times1\times2$
$=1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$ is equal to:
$\sin2(\theta+\phi)$
$\cos2(\theta+\phi)$
$\sin2(\theta-\phi)$
$\cos2(\theta-\phi)$
[Hint: Use
$\sin^2\text{A}-\sin^2\text{B}=\sin(\text{A+B})\sin(\text{A}-\text{B})$]If $\frac{(1-3\text{P})}{2},\frac{(1+4\text{P})}{3},\frac{(1+\text{P})}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is:
Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to:
Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is:
$\frac{1}{5}$
$\frac{4}{5}$
$\frac{1}{30}$
$\frac{5}{9}$