JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

Let

f(x)=\frac{\tan x}{x}

then

{\log _e}\left( {\mathop {\operatorname{Lt} }\limits_{x \to 0} {{\left( {[f(x)] + {x^2}} \right)}^{\frac{1}{{\{ f(x)\} }}}}} \right) =

________ Where [.] is G.I.F

\{\cdot\}

is fractional part function

View answer and explanation

Correct option: C

Option B (

\frac{1}{9}<I^2<\frac{1}{8}

) is correct because by analyzing the function within the integral and finding its bounds, the inequality for

I^2

is determined.

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