Let f(x)=tanx/x, then log(lim_{x tends to 0}([f(x)]+x²)^1/{f(x)}) is equal to, (where[.] denotes greatest integer function and{.} fractional part)=???

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$$\begin{gathered} \mathrm{This} \mathrm{is} \mathrm{a} \mathrm{very} \mathrm{conceptual} \mathrm{question}. \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{tanx}}{\mathrm{x}} \\ \mathrm{Now} \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{tanx}}{\mathrm{x}} \mathrm{is} \mathrm{slightly} \mathrm{greater} \mathrm{than} 1. \\ \mathrm{Hence} \underset{\mathrm{x}\to 0}{\lim } \left[\frac{\mathrm{tanx}}{\mathrm{x}}\right]=1 \mathrm{as} \left[\mathrm{x}\right]=1, \mathrm{x} \mathrm{is} \mathrm{slightly} \mathrm{greater} \mathrm{than} 1. \\ \mathrm{As} \left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ \mathrm{Hence} \\ \left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\mathrm{f}\left(\mathrm{x}\right)-\left[\mathrm{f}\left(\mathrm{x}\right)\right] \\ \Rightarrow \underset{\mathrm{x}\to 0}{\lim }\left\{\frac{\mathrm{tanx}}{\mathrm{x}}\right\}= \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{tanx}}{\mathrm{x}}-\underset{\mathrm{x}\to 0}{\lim } \left[\frac{\mathrm{tanx}}{\mathrm{x}}\right]=1-1=0 \\ \mathrm{Now} \\ \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } {\left(\left[\mathrm{f}\left(\mathrm{x}\right)\right]+{\mathrm{x}}^2\right)}^{\frac{1}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}} \\ =\underset{\mathrm{x}\to 0}{\lim } {\left(1+{\mathrm{x}}^2\right)}^{\frac{1}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}} \left[1^{\frac{1}{0}}=1^{\infty }\right] \\ \mathrm{This} \mathrm{is} \mathrm{nothing} \mathrm{but} 1^{\infty } \mathrm{form}. \\ \mathrm{Now} \mathrm{we} \mathrm{solve} \mathrm{it} \mathrm{in} \mathrm{following} \mathrm{manner}. \\ \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{ }{\left(\mathbf{f}\left(\mathbf{x}\right)\right)}^{\mathbf{g}\left(\mathbf{x}\right)}\mathbf{=}{\mathbf{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left(\mathbf{f}\left(\mathbf{x}\right)\mathbf{-}\mathbf{1}\right)\mathbf{g}\left(\mathbf{x}\right)}\mathbf{,}\mathbf{ }\mathbf{where}\mathbf{ }\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{and}\mathbf{ }\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{ }\mathbf{g}\left(\mathbf{x}\right)\mathbf{=}\mathbf{0} \\ \mathrm{Hence} \\ \underset{\mathrm{x}\to 0}{\lim } {\left(1+{\mathrm{x}}^2\right)}^{\frac{1}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}}={\mathrm{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}} \left(1+{\mathrm{x}}^2-1\right).\frac{1}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}}={\mathrm{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}} \frac{{\mathrm{x}}^2}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}} \\ \mathrm{Now} \mathrm{our} \mathrm{concern} \mathrm{is}: \\ \mathrm{M}=\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{\mathrm{f}\left(\mathrm{x}\right)-\left[\mathrm{f}\left(\mathrm{x}\right)\right]} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{{\displaystyle \frac{\mathrm{tanx}}{\mathrm{x}}}-1} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^3}{\mathrm{tanx}-\mathrm{x}} \\ \mathrm{As} \mathrm{it} \mathrm{is} \frac{0}{0} \mathrm{form}, \mathrm{hence} \mathrm{we} \mathrm{can} \mathrm{differentiate} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{b} \\ \mathrm{L}-\mathrm{Hopital} \mathrm{rule}. \\ \Rightarrow \mathrm{M}=\underset{\mathrm{x}\to 0}{\lim }\frac{3{\mathrm{x}}^2}{{\sec }^2\mathrm{x}-1}=3 \underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{{\displaystyle \frac{1}{{\cos }^2\mathrm{x}}}-1}=3\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2}{{\displaystyle \frac{1-{\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}}}} \\ =3\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{{\sin }^2\mathrm{x}}.{\cos }^2\mathrm{x} \\ =3\underset{\mathrm{x}\to 0}{\lim } {\left(\frac{\mathrm{x}}{\mathrm{sinx}}\right)}^2.{\cos }^2\mathrm{x} \\ =3 \times 1\times 1 \left[\mathrm{As} \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{x}}{\mathrm{sinx}}=1 \mathrm{and} \cos \left(0\right)=1\right] \\ =3 \\ \mathrm{Hence} \mathrm{M}=\underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}=3 \\ \Rightarrow {\mathrm{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}} \frac{{\mathrm{x}}^2}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}}={\mathrm{e}}^3 \\ \mathrm{Now} \\ \log \left[\underset{\mathrm{x}\to 0}{\lim } {\left(\left[\mathrm{f}\left(\mathrm{x}\right)\right]+{\mathrm{x}}^2\right)}^{\frac{1}{\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}}\right]=\log \left({\mathrm{e}}^3\right)=3 \left(\mathrm{Answer}\right) \left[\mathrm{As} {\log }_{\mathrm{e}}{\mathrm{e}}^{\mathrm{x}}=\mathrm{x}\right] \end{gathered}$$

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