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  •  Evaluate <img alt="\mathrm{\lim _{x \rightarrow \tan ^{-1} 3} \frac{\left[\tan ^2 x\right]-2[\tan x]-3}{\left[\tan ^2 x\right]-4[\tan x]+3}}" src="https://entrancecorner.oncodecogs.com/gif.la

 Evaluate \mathrm{\lim _{x \rightarrow \tan ^{-1} 3} \frac{\left[\tan ^2 x\right]-2[\tan x]-3}{\left[\tan ^2 x\right]-4[\tan x]+3}} where, \left [ \right ] denotes G.I.F.

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{3}


Option: 3

3


Option: 4

\frac{3}{2}


Answers (1)

best_answer

\mathrm{Let x=\tan ^{-1} 3+h }

So,
\begin{aligned} &\mathrm{ \tan x>3 \Rightarrow \tan ^2 x>9 }\\ &\mathrm{ {\left[\tan ^2 x\right]=9 \text { and }[\tan x]=3}} \\ &\mathrm{ {\left[\tan ^2 x\right]-4[\tan x]+3=9-4 \cdot 3+3=0}} \end{aligned}

Thus, denominator will become exact zero not approaching zero.

Hence not defined.

Thus, \mathrm{\tan x>3}  is not possible.

\begin{gathered} \mathrm{x=\tan ^{-1} 3-h \text { so, } \tan x<3 \Rightarrow \tan ^2 x<9 }\\ \mathrm{{\left[\tan ^2 x\right]=8 \text { and }[\tan x]=2} }\\ \mathrm{\lim _{x \rightarrow \tan ^{-1} 3} \frac{8-2 \cdot 2-3}{8-4 \cdot 2+3}=\frac{1}{3}} \end{gathered}
Now, we will discuss few problems based on expansion.
Expansion is a very powerful method to solve all limit problems.

Posted by

Devendra Khairwa

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