Question

solve!!!!!!!!!!!!!!!!!!!!!!!​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to \infty } {\bigg(\dfrac{x}{x + 3} \bigg) }^{3x}$

If we use direct substitution method, we get indeterminant form.

So, to evaluate this limit, Let assume that

$\rm :\longmapsto\:y = \displaystyle\lim_{x \to \infty } {\bigg(\dfrac{x}{x + 3} \bigg) }^{3x}$

On taking log both sides, we get

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty } \: log {\bigg(\dfrac{x}{x + 3} \bigg) }^{3x}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }3x \: log {\bigg(\dfrac{x}{x + 3} \bigg) }$

$\red{\bigg \{ \because \: log( {x}^{y} ) = y \: logx \bigg \}}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{ log(x) - log(x + 3) }{\dfrac{1}{3x} }$

$\red{\bigg \{ \because \: log\bigg(\dfrac{x}{y} \bigg) = logx - logy \bigg \}}$

Using L - Hospital Rule, we have

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{\dfrac{d}{dx} (log(x) - log(x + 3) )}{\dfrac{d}{dx} \bigg(\dfrac{1}{3x} \bigg) }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{\dfrac{1}{x} - \dfrac{1}{x + 3} }{\dfrac{ - 1}{3 {x}^{2} } }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{\dfrac{x + 3 - x}{x(x + 3)}}{\dfrac{ - 1}{3 {x}^{2} } }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{\dfrac{3}{(x + 3)}}{\dfrac{ - 1}{3 {x} } }$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{ - 9x}{x + 3}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{ - 9x}{x(1 + \dfrac{3}{x} )}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{x \to \infty }\dfrac{ - 9}{1 + \dfrac{3}{x}}$

$\rm :\longmapsto\:logy = \dfrac{ - 9}{1 + 0}$

$\rm :\longmapsto\:logy = - 9$

$\rm :\longmapsto\:y = {e}^{ - 9}$

$\rm :\longmapsto\:y = \dfrac{1}{ {e}^{9} }$

Hence,

$\bf :\longmapsto\:\displaystyle\lim_{x \to \infty } \bf \: {\bigg(\dfrac{x}{x + 3} \bigg) }^{3x} = \dfrac{1}{ {e}^{9} }$

Additional Information :-

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{sinx}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{tanx}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{ {sin}^{ - 1} x}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{ {tan}^{ - 1} x}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{log(1 + x)}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{ {e}^{x} - 1}{x} = 1}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to 0 } \bf \: \dfrac{ {a}^{x} - 1}{x} = loga}}}$

$\blue{\boxed{ \bf{\displaystyle\lim_{x \to \infty } \bf \: \bigg(1 + \dfrac{1}{x} \bigg)^{x} = e}}}$