Question

$\underset{x\to 0}{\lim }{\left(\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\right)}^{\frac{1}{x}}=$

• A. $e$
• B. $e^2$
• C. $2$
• D. $1$

Answer 1

The correct option is B

$e^2$

Explanation for Correct Answer:

Evaluating the given function:

Given,

$$\begin{gathered} limi{t}_{x\to 0}{\left(\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\right)}^{\frac{1}{x}}=limi{t}_{x\to 0}{\left(\frac{\tan \left(\frac{\mathrm{\pi }}{4}\right)+\tan \left(x\right)}{1-\tan \left(\frac{\mathrm{\pi }}{4}\right)\tan \left(x\right)}\right)}^{\frac{1}{x}}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\left(A+B\right)\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\left(A\right)\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\left(B\right)}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\left(A\right)\mathbf{t}\mathbf{a}\mathbf{n}\left(B\right)}\mathbf{]} \\ =limi{t}_{x\to 0}{\left(\frac{1+\tan \left(x\right)+\tan \left(x\right)-\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}}\mathbf{\left[}\mathbf{\text{Add and subtract}}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\text{numerator}}\mathbf{\right]} \\ =limi{t}_{x\to 0}{\left(\frac{\left(1-\tan \left(x\right)\right)+2\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}} \\ =limi{t}_{x\to 0}{\left(\frac{1-\tan \left(x\right)}{1-\tan \left(x\right)}+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}} \\ =limi{t}_{x\to 0}{\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}} \\ =1^{\infty }\text{form} \end{gathered}$$

If $limi{t}_{x\to a}f{\left(x\right)}^{g\left(x\right)}=1^{\infty },$then $limi{t}_{x\to a}f{\left(x\right)}^{g\left(x\right)}={e^{limi{t}_{x\to a}\left(f\left(x\right)-1\right)}}^{g\left(x\right)}$

$limi{t}_{x\to 0}{\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}}=e^{limi{t}_{x\to 0}\frac{1}{x}.\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}-1\right)}....\left(i\right)$

First, we solve

$$\begin{gathered} limi{t}_{x\to 0}\frac{1}{x}.\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}-1\right)=2limi{t}_{x\to 0}\left(\frac{\tan \left(x\right)}{x}\right).\left(\frac{1}{1-\tan \left(x\right)}\right) \\ =2limi{t}_{x\to 0}\left(\frac{\tan \left(x\right)}{x}\right).limi{t}_{x\to 0}\left(\frac{1}{1-\tan \left(x\right)}\right)\left[\because {\mathrm{limit}}_{x\to 0}\left(\frac{\tan \left(x\right)}{x}\right)=1\right]\mathbf{} \\ =2\left(1\right).\left(\frac{1}{1-\tan \left(0\right)}\right) \\ =2\mathbf{}\left[\because \tan \left(0\right)=1\right] \end{gathered}$$

Substitute in equation $\left(i\right)$

$$\begin{gathered} limi{t}_{x\to 0}{\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}\right)}^{\frac{1}{x}}=e^{limi{t}_{x\to 0}\frac{1}{x}.\left(1+\frac{2\tan \left(x\right)}{1-\tan \left(x\right)}-1\right)} \\ =e^2\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{limit}}_{\mathbf{x}\mathbf{\to }\mathbf{0}}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{2}\mathbf{tan}\left(x\right)}{\mathbf{1}\mathbf{-}\mathbf{tan}\left(x\right)}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{]} \end{gathered}$$

Hence, the correct answer is option (B).