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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}$ At $\text{x}=\frac{\pi}{2}$, the value of the given function takes the form $\frac{0}{0}$. Now, put $\text{x}-\frac{\pi}{2}=\text{y}$ so that $\text{x}\rightarrow\frac{\pi}{2},\text{y}\rightarrow0$. $\therefore\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}=\lim\limits_{\text{y}\rightarrow0}\frac{\tan2\big(\text{y}+\frac{\pi}{2}\big)}{\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\tan(\pi+2\text{y})}{\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\tan2\text{y}}{\text{y}}$ $[\tan(\pi+2\text{y})=\tan2\text{y}]$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sin2\text{y}}{\text{y}\cos2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\bigg(\frac{\sin2\text{y}}{2\text{y}}\times\frac{2}{\cos2\text{y}}\bigg)$ $=\bigg(\lim\limits_{2\text{y}\rightarrow0}\frac{\sin2\text{y}}{2\text{y}}\bigg)\times\lim\limits_{\text{y}\rightarrow0}\bigg(\frac{2}{\cos2\text{y}}\bigg)$ $[\text{y}\rightarrow0\Rightarrow2\text{y}\rightarrow0]$ $=1\times\frac{2}{\cos0}$ $\big[=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\big]$$=1\times\frac{2}{1}$
$=2$

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