Lim_x → 11 x^2/sin π x =lim_x → 1(1 x)(1 + x)/sin π x ⇒ x → 1, x 1 → 0, let x 1 = y ⇒ y → 0 =lim_y → 0( y)(1 + y + 1)/sin π (y+1) = lim_y → 0y(y + 2)/si ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

lim x → 11-x2...

Question

${lim}_{x \to 1} \frac{1 - x^{2}}{s i n π x}$

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Solution

${lim}_{x \to 1} \frac{1 - x^{2}}{s i n π x}$

$= {lim}_{x \to 1} \frac{(1 - x) (1 + x)}{s i n π x}$

$\Rightarrow x \to 1, x - 1 \to 0,$ let $x - 1 = y \Rightarrow y \to 0$

$= {lim}_{y \to 0} \frac{(- y) (1 + y + 1)}{s i n π (y + 1)}$

$= - {lim}_{y \to 0} \frac{y (y + 2)}{s i n (π y + π)}$

$= - {lim}_{y \to 0} \frac{y (y + 2)}{- s i n π y}$

$= {lim}_{y \to 0} \frac{y (y + 2)}{\frac{s i n π y}{2}}$

$= \frac{{lim}_{y \to 0} y (y + 2)}{{lim}_{y \to 0} \frac{s i n π y}{π y} \times π y} = \frac{2}{π}$

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limx→11−x2sinπx

limx→11−x2sinπx =limx→1(1−x)(1+x)sinπx ⇒x→1,x−1→0, let x−1=y⇒y→0 =limy→0(−y)(1+y+1)sinπ(y+1) =−limy→0y(y+2)sin(πy+π) =−limy→0y(y+2)−sinπy =limy→0y(y+2)sinπy2 =limy→0y(y+2)limy→0sin πyπy×πy=2π

${lim}_{x \to 1} \frac{1 - x^{2}}{s i n π x}$