Question

\lim _ { x \rightarrow 1 } \frac { 1 - x ^ { 2 } } { \sin \pi x }​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to 1}\rm \frac{1 - {x}^{2} }{sin \: \pi \: x}$

If we substitute directly x = 1, we get

$\rm \: = \: \dfrac{1 - 1}{sin\pi}$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form

Now, Consider

$\rm \: \displaystyle\lim_{x \to 1}\rm \frac{1 - {x}^{2} }{sin \: \pi \: x}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \frac{(1 + x)(1 - x)}{sin \: \pi \: x}$

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm (x + 1) \times \displaystyle\lim_{x \to 1}\rm \frac{1 - x}{sin \: \pi \: x}$

$\rm \: = \: 2 \displaystyle\lim_{x \to 1}\rm \frac{1 - x}{sin \: \pi \: x}$

To evaluate this limit, we use method of Substitution,

So, substitute

$\rm \: x = 1 - h, \: \: \: as \: x \: \to \: 1, \: \: so \: h \: \to \: 0$

So, above expression can be rewritten as

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \dfrac{1 - (1 - h)}{sin\pi \: (1 - h)}$

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \dfrac{1 - 1 + h}{sin(\pi - \pi \: h)}$

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \dfrac{h}{sin \: \pi \: h}$

can be rewritten as

$\rm \: = \: 2\displaystyle\lim_{h \to 0}\rm \dfrac{\pi \: h}{sin \: \pi \: h} \times \frac{1}{\pi}$

can be further rewritten as

$\rm \: = \: 2\displaystyle\lim_{\pi \: h \to 0}\rm \dfrac{\pi \: h}{sin \: \pi \: h} \times \frac{1}{\pi}$

We know,

$\color{green}\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: \: }}$

So, using this result, we get

$\rm \: = \: 2 \times 1 \times \dfrac{1}{\pi}$

$\rm \: = \: \dfrac{2}{\pi}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\lim_{x \to 1}\rm \: \: \frac{1 - {x}^{2} }{sin \: \pi \: x} = \frac{2}{\pi} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\implies \lim_{x \to \: 1} \: \frac{1 - x {}^{2} }{ \sin \pi \: x} \\ \\ put \: limit \\ \\ \implies \: \frac{1 - {1}^{2} }{ \sin \pi \: } \\ \\ \implies \: \frac{0}{0} \\ \\ indeterminant \: form \: \\ so \: we \: use \: l \: hospital \\ \\ \implies \: \lim_{x \to \: 1} \: \frac{d}{dx} \frac{1 - {x}^{2} }{ \frac{d}{dx} \sin \pi \: x \: . \frac{d}{dx} \pi \: x} \\ \\ \implies \: \lim_{x \to \: 1} \: \frac{ - 2x}{ \cos \pi \: x \: . \pi} \\ \\ \implies \: \lim_{x \to \: 1} \: \: \frac{ - 2x}{ \pi \: . \cos \pi \: x} \\ \\ put \: limit \\ \\ \implies \: \frac{ - 2(1)}{ \pi \: cos \pi} \\ \\ \implies \: \frac{ \cancel{-} \: \: 2}{ \pi \: ( \cancel{-} \: \: 1)} \\ \\ \implies \boxed{ \red{ \frac{2}{ \pi}} }$

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thanks