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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The vertex of the parabola y2 - 4y - x + 3 = 0 is:
$\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}$ is equal to:
$2$
$\frac{3}{2}$
$-\frac{3}{2}$
$1$