Question

$\underset{x\to 1}{\lim }\frac{\tan (x^2-1)}{x-1}$ is equal to _____________________.

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 1\end{array}\frac{\tan \left(x^2-1\right)}{x-1} \\ =\begin{array}{c}\lim \\ x\to 1\end{array}\frac{\tan \left(x^2-1\right)}{x-1}\times \frac{x+1}{x+1} \\ =\begin{array}{c}\lim \\ x\to 1\end{array}\frac{\tan \left(x^2-1\right)}{\left(x^2-1\right)}\times x+1 \\ \mathrm{i}.\mathrm{e}.\left(\begin{array}{c}\lim \\ x\to 1\end{array}\frac{\tan \left(x^2-1\right)}{\left(x^2-1\right)}\right)\left(\begin{array}{c}\lim \\ x\to 1\end{array}x+1\right)\left(\mathrm{since}\begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{\tan \theta }{\theta }=1\right) \\ \mathrm{i}.\mathrm{e}.1\times \left(1+1\right)=2 \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 1\end{array}\frac{\tan \left(x^2-1\right)}{x-1}=2 \end{gathered}$$