Question

Find the value of given limit $ \mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^2}}}{{\sin \pi x}} $

Answer 1

Hint: For limit problems in which the limit of the function is not zero, we will make the limit of the function first zero. This can be done by using substitution of x = a + h with $ h \to 0 $ and then simplifying the limit of the function so formed after substitution by using standard limit formulas.
  $ \sin \left( {\pi + x} \right) = - \sin x $ and \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]

Complete step-by-step answer:
In the given limit problem we see that the limit of function is not zero.
Therefore, we first make the limit to zero by doing substitution.

Let $ x = 1 + h $ where $ h \to 0 $
Substituting above value of ‘x’ in given limit function we have
  $ \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - {{\left( {1 + h} \right)}^2}}}{{\sin \pi \left( {1 + h} \right)}} $
\[
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - {{\left( {1 + h} \right)}^2}}}{{\sin \left( {\pi + \pi h} \right)}} \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - {{\left( {1 + h} \right)}^2}}}{{ - \sin \left( {\pi h} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \sin \left( {\pi + \theta } \right) = - \sin \theta \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - \left( {1 + {h^2} + 2h} \right)}}{{ - \sin \left( {\pi h} \right)}} \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - 1 + {h^2} + 2h}}{{ - \sin \left( {\pi h} \right)}} \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{ - {h^2} - 2h}}{{ - \sin \left( {\pi h} \right)}}\, \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{ - h\left( {h + 2} \right)}}{{ - \sin \left( {\pi h} \right)}}\, \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {h + 2} \right)}}{{\sin \left( {\pi h} \right)}}\, \\
 \]

Multiply and divide by $ \pi h $ in the denominator of the above equation.
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {h + 2} \right)}}{{\dfrac{{\sin \left( {\pi h} \right)}}{{\pi h}} \times \pi h}}\,\]
\[
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {h + 2} \right)}}{{\pi h}}\, \\
   \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {h + 2} \right)}}{\pi }\, \\
 \]
Substituting limit $ h \to 0 $ we have
  $ \dfrac{2}{\pi } $
Hence, from above we see that limit of the function $ \mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {x^2}}}{{\sin \pi x}} $ is $ \dfrac{2}{\pi } $ .

Note: For all limits problems if limit is not zero then first make its limit zero by substitution and then using standard formulas of limits like $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1,\,\,\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\,\,and\,\,\mathop {\lim }\limits_{x \to 0} \cos x = 1 $ to evaluate limit of the function so formed after substitution.