Question

lim x →a {1-cos(x-a)}²/(x-a)⁴​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to a}\rm \frac{ {\bigg(1 - cos(x - a) \bigg) }^{2} }{ {(x - a)}^{4} }$

If we substitute directly x = a, we get

$\rm \:  =  \: \dfrac{ {\bigg(1 - cos(a - a)\bigg) }^{2} }{ {(a - a)}^{4} }$

$\rm \:  =  \: \dfrac{ {\bigg(1 - cos0\bigg) }^{2} }{ {0}^{4} }$

$\rm \:  =  \: \dfrac{ {\bigg(1 -1\bigg) }^{2} }{0}$

$\rm \:  =  \: \dfrac{ {\bigg(0\bigg) }^{2} }{0}$

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

So, as to evaluate

$\rm :\longmapsto\:\displaystyle\lim_{x \to a}\rm \frac{ {\bigg(1 - cos(x - a) \bigg) }^{2} }{ {(x - a)}^{4} }$

we use method of Substitution,

$\red{ \boxed{ \sf{ \:Put \: x = a + h, \: \: as \: x \: \to \: a, \: \: so \: h \: \to \: 0}}}$

So, above expression can be rewritten as

$\rm \: = \:\displaystyle\lim_{h \to 0}\rm \frac{ {\bigg(1 - cos(a + h - a) \bigg) }^{2} }{ {(a + h - a)}^{4} }$

$\rm \: = \:\displaystyle\lim_{h \to 0}\rm \frac{ {\bigg(1 - cosh \bigg) }^{2} }{ {h}^{4} }$

$\rm \: = \:\displaystyle\lim_{h \to 0}\rm \frac{ {\bigg(2 {sin}^{2}\dfrac{h}{2} \bigg) }^{2} }{ {h}^{4} }$

$\rm \: = \:\displaystyle\lim_{h \to 0}\rm \frac{ {4 {sin}^{4}\dfrac{h}{2}}}{ {h}^{4} }$

$\rm \:  = 4 \: \displaystyle\lim_{h\to 0}\rm {\bigg[\dfrac{sin\dfrac{h}{2} }{h} \bigg]}^{4}$

$\rm \:  = 4 \: \displaystyle\lim_{h\to 0}\rm {\bigg[\dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2} \times 2} \bigg]}^{4}$

$\rm \:  = \dfrac{4}{16} \: \displaystyle\lim_{h\to 0}\rm {\bigg[\dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2}} \bigg]}^{4}$

We know,

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}}$

So, using this, we get

$\rm \:  = \dfrac{1}{4} \: \times {1}^{4}$

$\rm \:  = \dfrac{1}{4}$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to a}\rm \frac{ {\bigg(1 - cos(x - a) \bigg) }^{2} }{ {(x - a)}^{4} } = \frac{1}{4}}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}}$