Question
$\lim\limits_{\text{x} \rightarrow0}\frac{\text{cosec}-\cot\text{x}}{\text{x}}$ is equal to:
-
$-\frac{1}{2}$
-
$1$
-
$\frac{1}{2}$
-
$-1$
$\lim\limits_{\text{x} \rightarrow0}\frac{\text{cosec}-\cot\text{x}}{\text{x}}$ is equal to:
$-\frac{1}{2}$
$1$
$\frac{1}{2}$
$-1$
Solution:
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\text{cosec}\text{x}-\cot\text{x}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}{\text{x}}$
$ =\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}=\frac{2\sin^{2}\frac{\text{x}}{2}}{\text{x}\cdot\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\sin\frac{\text{x}}{2}}{\text{x}\cos\frac{\text{x}}{2}} =\frac{\tan\frac{\text{x}}{2}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\frac{\text{x}}{2}}{2\times\frac{\text{x}}{2}}$
$=\frac{1}{2}\times1=\frac{1}{2}$
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Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals: