MCQ
Choose the correct answer.
$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
- A3
- B1
- C0
- D2
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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