Question

find the limit .of this ​

Answer 1

$\rm :\longmapsto\:\displaystyle\lim_{x \to \: 0}\sf \: \dfrac{1 - {cos}^{3}x }{sin3x \: sin5x}$

$\rm \: = \: \: \sf \: \dfrac{1 - {cos}^{3}0 }{sin(3 \times 0) \: sin(5 \times 0)}$

$\rm \: = \: \: \sf \: \dfrac{1 - 1}{0 \times 0}$

$\rm \: = \: \: \sf \: \dfrac{0}{0} \: which \: is \: meaningless$

Thus,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \: 0}\sf \: \dfrac{1 - {cos}^{3}x }{sin3x \: sin5x}$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf (1 - {cos}^{3}x) \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{1}{sin3x} \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{1}{sin5x}$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf (1 - {cos}^{3}x) \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{3x}{sin3x} \times \dfrac{1}{3x} \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{5x}{sin5x} \times \dfrac{1}{5x}$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf (1 - {cos}^{3}x) \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{1}{3x} \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{1}{5x}$

$\red{\bigg \{ \because \:\displaystyle\lim_{x \to \: 0}\sf \dfrac{sinx}{x} = 1\bigg \}}$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf \dfrac{1 - {cos}^{3} x}{15 {x}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf \dfrac{(1 - {cos} x)(1 + cosx + {cos}^{2}x) }{15 {x}^{2} }$

$\green{\boxed{ \because \: \bf \: {x}^{3}- {y}^{3} = (x - y)( {x}^{2} + xy+{y}^{2}}}$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf \dfrac{(1 - {cos} x)}{15 {x}^{2} } \times \displaystyle\lim_{x \to \: 0}\sf (1 + cosx + {cos}^{2}x)$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf \dfrac{2 {sin}^{2}\dfrac{x}{2}}{15 {x}^{2} } \times (1 + cos0 + {cos}^{2}0)$

$\rm \: = \: \: \displaystyle\lim_{x \to \: 0}\sf \dfrac{2 {sin}\dfrac{x}{2} \times sin\dfrac{x}{2} }{15 {x}^{2} } \times (1 + 1 + 1)$

$\rm \: = \: \: \dfrac{2}{5} \displaystyle\lim_{x \to \: 0}\sf \dfrac{ {sin}\dfrac{x}{2} \times sin\dfrac{x}{2} }{{x}^{2} }$

$\rm \: = \: \: \dfrac{2}{5} \displaystyle\lim_{x \to \: 0}\sf \dfrac{ {sin}\dfrac{x}{2} \times sin\dfrac{x}{2} }{{x} \times x}$

$\rm \: = \: \: \dfrac{2}{5} \displaystyle\lim_{x \to \: 0}\sf \dfrac{sin\dfrac{x}{2} }{\dfrac{x}{2} \times 2} \times \displaystyle\lim_{x \to \: 0}\sf \dfrac{sin\dfrac{x}{2} }{\dfrac{x}{2} \times 2}$

$\red{\bigg \{ \because \:\displaystyle\lim_{x \to \: 0}\sf \dfrac{sinx}{x} = 1\bigg \}}$

$\rm \: = \: \: \dfrac{2}{5} \times \dfrac{1}{2} \times \dfrac{1}{2}$

$\rm \: = \: \: \dfrac{1}{10}$

Additional Information :-

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

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

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

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

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

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

$\green{\boxed{ \bf \:\displaystyle\lim_{x \to \: 0}\sf \dfrac{ {a}^{x} - 1}{x} = log(a) }}$