Question
$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
- 3
- 1
- 0
- 2
$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
Solution:
Given $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\tan\text{x}-1}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{(\tan\text{x}+1)(\tan\text{x}-1)}{(\tan\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)=\tan\frac{\pi}{4}+1$
$=1+1=2$
$=2$
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If A and B are coefficient of xn in the expansions of (1 + x)2n and (1 + x)2n – 1 respectively, then $\frac{\text{A}}{\text{B}}$ equals:
$1.$
$2.$
$\frac{1}{2}.$
$\frac{1}{\text{n}}.$
Hint:
$\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=2$