If g(x) = \underset{t \rightarrow 0}{Lim} \textrm{ }\textrm{ } tan \left(\right. \frac{e^{2 tx} - 2 \textrm{ } tx - 1}{2 t^{2}} \left.\right) and \textrm{ }\textrm{ } L = \underset{x \rightarrow 0}{Lim} \textrm{ }\textrm{ } \frac{g^{2} \left(\right. x \left.\right) - x^{4}}{x^{8}} then [Note : {y} and [y] denote fractional part function and greatest integer function of y respectively.]

Engineering

Mathematics

Trigonometric Limits

Question

If g(x) = $\underset{t \to 0}{Lim} \tan (\frac{e^{2 tx} - 2 tx - 1}{2 t^{2}}) and L = \underset{x \to 0}{Lim} \frac{g^{2} (x) - x^{4}}{x^{8}}$ then

[Note : {y} and [y] denote fractional part function and greatest integer function of y respectively.]

$\underset{x \to 0}{Lim} \frac{g (L \cdot x)}{x^{2}} = \frac{25}{9}$

[L] = 1

$\underset{x \to 0}{Lim} \frac{g (L \cdot x)}{x^{2}} = \frac{4}{9}$

${L} = \frac{2}{3}$

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$g (x) = \underset{t \to 0}{Lim} \tan (\frac{e^{2 tx} - 2 tx - 1}{4 t^{2} x^{2}} \cdot 2 x^{2}) = \tan (x^{2})$

$L = \underset{x \to 0}{Lim} \frac{\tan^{2} (x^{2}) - x^{4}}{x^{8}}$

$= \underset{x \to 0}{Lim} \frac{({tanx}^{2} + x^{2})}{x^{2}} \cdot \frac{({tanx}^{2} - x^{2})}{x^{6}}$

$= 2 \times \frac{1}{3} = \frac{2}{3}$

Clearly, [L] = 0, {L} = $\frac{2}{3}$

$\underset{x \to 0}{Lim} \frac{g (L \cdot x)}{x^{2}} = \underset{x \to 0}{Lim} \frac{g (\frac{2}{3} x)}{x^{2}} = \underset{x \to 0}{Lim} \frac{\tan (\frac{4}{9} x^{2})}{x^{2}} = \frac{4}{9}$