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lim_{x\rightarrow 1} sin(x-1)/x^{2}-1

 

Answers (1)

Substitute x=1 to test this function.

\begin{array}{l} \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{2}-1}=\frac{\sin (1-1)}{(1)^{2}-1} \\ \Longrightarrow \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{2}-1}=\frac{\sin (0)}{1-1} \\ \Longrightarrow \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{2}-1}=\frac{0}{0} \end{array}

In numerator, the angle is x−1 and it can be obtained in denominator as well by expressing x2−1 as two multiplying factors in which x−1 is one of them.

\Longrightarrow \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{2}-1}=\lim _{x \rightarrow 1} \frac{\sin (x-1)}{(x-1)(x+1)}

=\lim _{x-1 \rightarrow 0}\left[\frac{\sin (x-1)}{(x-1)}\right] \times \lim _{x \rightarrow 1}\left[\frac{1}{(x+1)}\right]

Therefore, the value of first multiplying factor is 1 and substitute x=1 in the second multiplying factor to obtain the required solution.

\begin{array}{l} =1 \times \frac{1}{1+1} \\\\ =1 \times \frac{1}{2} \\\\ \therefore \quad \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{2}-1}=\frac{1}{2} \end{array}

 

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Satyajeet Kumar

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