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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
As x → 1, then x - 1 → 0 let x - 1 = y
$=\lim\limits_{{\text{x}-{1\rightarrow0}}}\frac{(\text{x}-1)}{\text{x}\times\sin\pi(\text{x}-1)}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{1}{\frac{(\text{y}+1)\sin(\pi\text{y})}{\text{y}}}$
$=\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow0}(\text{y}+1)\Big)\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\pi\text{y}}{\text{y}\times\pi}\times\pi\Big)}$
$=\frac{1}{(1)(1\times\pi)}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{\pi}$

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