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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
⇒ x → 1, then x - 1 →0, let x - 1 = y
$=\lim\limits_{(\text{x}-1)\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin2\pi\text{ x}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{-\text{y}(1+\text{y}+1)}{\sin2\pi(\text{y}+1)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin(2\pi\text {y}+2\pi)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin2\pi\text {y}}$
$=-\lim\limits_{\text{y}\rightarrow{0}}(\text{y}+2)\times\frac{\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\sin\frac{2\pi\text {y}}{\text {y}\times2\pi}\Big)\times2\pi\text {y}}$
$=-2\times\frac{1}{1\times2\pi}$
$=-\frac{1}{\pi}$

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