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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}\times\pi}{180}}{\text{x}}$ $\Big[\because\ 1^\circ=\frac{\pi}{180}\text{ radians}\Big]$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}\times\frac{\pi}{180}}\times\frac{\pi}{180}$
$=\frac{\pi}{180}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\pi\text{x}}{180}}{\frac{\text{x}\pi}{180}}$
$=\frac{\pi}{180}\times1=\frac{\pi}{180}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{\pi}{180}$

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