Question

Solve the given limit.
$\displaystyle\lim_{x\to 0}\dfrac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}$

Dont use L'Hôpital's rule.

Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x\to 0}\dfrac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3}$

We know,

$\boxed{ \rm{ \:ln(xy) = ln(x) + ln(y) \: }} \\ \\ \boxed{ \rm{ \:ln \frac{x}{y} = ln(x) - ln(y) \: \: }} \\ \\ \boxed{ \rm{ \:ln( {x}^{y}) \: = \: yln(x) \: \: }}$

So, using these results, we get

$\rm \: = \displaystyle\lim_{x\to 0} \frac{ln(1 - 3x) + 3ln(1 + x) - ln(1 + 3x) - 3ln(1 - x)}{ {x}^{3} }$

We know,

$\rm \: ln(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + - - -$

So,

$\rm \: ln(1 - x) = - x - \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} - - -$

and

$\rm \: ln(1 - 3x) = - 3x - \dfrac{9{x}^{2} }{2} - \dfrac{27{x}^{3} }{3} - \dfrac{ 81{x}^{4} }{4} - - -$

and

$\rm \: ln(1 + 3x) = 3x - \dfrac{9{x}^{2} }{2} + \dfrac{27{x}^{3} }{3} - \dfrac{ 81{x}^{4} }{4} + - - -$

So,

$\rm \: ln(1 - 3x) - ln(1 + 3x) = - 6x - 18 {x}^{3} -\frac{486 {x}^{5} }{5} + - - -$

and

$\rm \: ln(1 + x) - ln(1 - x) = 2x + \dfrac{2 {x}^{3} }{3} + \dfrac{ {2x}^{5} }{5} + - - -$

So,

$\rm \:3 ln(1 + x) - 3ln(1 - x) = 6x + {2x}^{3} + \dfrac{ {6x}^{5} }{5} + - - -$

So, it means

$\rm \: ln(1 - 3x) - ln(1 + 3x) + 3ln(1 + x) - 3ln(1 - x)$

$\rm \: = \: - 16 {x}^{3} - \dfrac{480{x}^{5} }{5} + - - -$

$\rm \: = \: - 16 {x}^{3} - {96x}^{5} + - - -$

On substituting this value in

$\rm \: = \displaystyle\lim_{x\to 0} \frac{ln(1 - 3x) + 3ln(1 + x) - ln(1 + 3x) - 3ln(1 - x)}{ {x}^{3} }$

we get

$\rm \: = \: \displaystyle\lim_{x\to 0} \frac{ - 16 {x}^{3} - 96 {x}^{5} - - - - }{ {x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x\to 0} \frac{ {x}^{3}[ - 16 - 96 {x}^{2} - - - - ]}{ {x}^{3} }$

$\rm \: = \: \displaystyle\lim_{x\to 0}[ - 16 - {96x}^{2} - - - ]$

$\rm \: = \: - \: 16$

Hence,

$\\ \rm\implies \:\boxed{ \rm{ \:\displaystyle\lim_{x\to 0}\dfrac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{x^3} \: = \: - \: 16 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3} }{3!} + - - - }\\ \\ \bigstar \: \bf{ {e}^{ - x} = 1 - x + \dfrac{ {x}^{2} }{2!} - \dfrac{ {x}^{3} }{3!} + - - - }\\ \\ \bigstar \: \bf{sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} + - - - }\\ \\ \bigstar \: \bf{cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} + - - - }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$