Question

$limi{t}_{x\to {\tan }^{-1}\left(3\right)}\frac{{\tan }^2\left(x\right)-2\tan \left(x\right)-3}{{\tan }^2\left(x\right)-4\tan \left(x\right)+3}=$

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is B

$2$

Explanation for correct answer:

Calculating the value:

$$\begin{gathered} {\mathrm{limit}}_{x\to {\tan }^{-1}\left(3\right)}\frac{{\tan }^2\left(x\right)-2\tan \left(x\right)-3}{{\tan }^2\left(x\right)-4\tan \left(x\right)+3}={\mathrm{limit}}_{x\to {\tan }^{-1}\left(3\right)}\frac{{\tan }^2\left(x\right)+\tan \left(x\right)-3\tan \left(x\right)-3}{{\tan }^2\left(x\right)-\tan \left(x\right)-3\tan \left(x\right)+3}\left[\because -2\tan \left(x\right)=-3\tan \left(x\right)+\tan \left(x\right)\right] \\ ={\mathrm{limit}}_{x\to {\tan }^{-1}\left(3\right)}\frac{\tan \left(x\right)\left(\tan \left(x\right)+1\right)-3\left(\tan \left(x\right)+1\right)}{\tan \left(x\right)\left(\tan \left(x\right)-1\right)-3\left(\tan \left(x\right)-1\right)} \\ ={\mathrm{limit}}_{x\to {\tan }^{-1}\left(3\right)}\frac{\left(\tan \left(x\right)+1\right)\left(\tan \left(x\right)-3\right)}{\left(\tan \left(x\right)-1\right)\left(\tan \left(x\right)-3\right)} \\ =\frac{\left(\tan \left({\tan }^{-1}\left(3\right)\right)+1\right)}{\left(\tan \left({\tan }^{-1}\left(3\right)\right)-1\right)} \\ =\frac{3+1}{3-1}\left[\because \tan \left({\tan }^{-1}\left(x\right)\right)=x\right] \\ =2 \end{gathered}$$

Hence, option (B) is the correct answer.