Question

Evaluate the given expression: \[{{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left( 1-\cos \dfrac{{{x}^{2}}}{2}-\cos \dfrac{{{x}^{2}}}{4}+\cos \dfrac{{{x}^{2}}}{2}\cos \dfrac{{{x}^{2}}}{4} \right)\] .

Answer 1

Hint: First of all, transform the expression \[{{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left( 1-\cos \dfrac{{{x}^{2}}}{2}-\cos \dfrac{{{x}^{2}}}{4}+\cos \dfrac{{{x}^{2}}}{2}\cos \dfrac{{{x}^{2}}}{4} \right)\] as
\[{{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left\{ \left( 1-\cos \dfrac{{{x}^{2}}}{2} \right)-\cos \dfrac{{{x}^{2}}}{4}\left( 1-\cos \dfrac{{{x}^{2}}}{2} \right) \right\}\] . Now, take the term \[\left( 1-\cos \dfrac{{{x}^{2}}}{2} \right)\] from the whole and simplify it. We know the formula, \[1-\cos 2\theta =2{{\sin }^{2}}\theta \] . Replace \[\theta \] by \[\dfrac{{{x}^{2}}}{4}\] in this formula and obtain one equation. Similarly, replace \[\theta \] by \[\dfrac{{{x}^{2}}}{8}\] in the formula and get other equation. Now, using these two equations, transform the expression. Now, break the term \[{{x}^{8}}\] as the product of \[{{x}^{4}}\] and \[{{x}^{4}}\] . Transform the expression and use the formula, \[{{\lim }_{x\to 0}}\dfrac{\sin x}{x}=1\] to simplify it further.

Complete step by step answer:
According to the question, we have the expression,
\[{{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left( 1-\cos \dfrac{{{x}^{2}}}{2}-\cos \dfrac{{{x}^{2}}}{4}+\cos \dfrac{{{x}^{2}}}{2}\cos \dfrac{{{x}^{2}}}{4} \right)\] ………………………………(1)
We can see that we don’t have any direct formula through which the given expression can be simplified and solved. So, we have to simplify the given expression into a simpler dorm.
Now, simplifying equation (1), we get
\[={{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left( 1-\cos \dfrac{{{x}^{2}}}{2}-\cos \dfrac{{{x}^{2}}}{4}+\cos \dfrac{{{x}^{2}}}{2}\cos \dfrac{{{x}^{2}}}{4} \right)\]
\[={{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left\{ \left( 1-\cos \dfrac{{{x}^{2}}}{2} \right)-\cos \dfrac{{{x}^{2}}}{4}\left( 1-\cos \dfrac{{{x}^{2}}}{2} \right) \right\}\] …………………………………………(2)
Now, taking the term \[\left( 1-\cos \dfrac{{{x}^{2}}}{2} \right)\] as common in equation (2), we get
\[={{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left\{ \left( 1-\cos \dfrac{{{x}^{2}}}{2} \right)\left( 1-\cos \dfrac{{{x}^{2}}}{4} \right) \right\}\] ……………………………………………(3)
We know the formula, \[1-\cos 2\theta =2{{\sin }^{2}}\theta \] ………………………………...(4)
Now, replacing \[\theta \] by \[\dfrac{{{x}^{2}}}{4}\] in equation (4), we get
\[\Rightarrow 1-\cos 2\times \dfrac{{{x}^{2}}}{4}=2{{\sin }^{2}}\dfrac{{{x}^{2}}}{4}\]
\[\Rightarrow 1-\cos \dfrac{{{x}^{2}}}{2}=2{{\sin }^{2}}\dfrac{{{x}^{2}}}{4}\] ……………………………….(5)
Now, replacing \[\theta \] by \[\dfrac{{{x}^{2}}}{8}\] in equation (4), we get
\[\Rightarrow 1-\cos 2\times \dfrac{{{x}^{2}}}{8}=2{{\sin }^{2}}\dfrac{{{x}^{2}}}{8}\]
\[\Rightarrow 1-\cos \dfrac{{{x}^{2}}}{4}=2{{\sin }^{2}}\dfrac{{{x}^{2}}}{8}\] ……………………………….(6)
Now, substituting equation (5) and equation (6) in equation (3), we get
\[={{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left\{ \left( 2{{\sin }^{2}}\dfrac{{{x}^{2}}}{4} \right)\left( 2{{\sin }^{2}}\dfrac{{{x}^{2}}}{8} \right) \right\}\] ……………………………..(7)
We need to make the term \[{{x}^{8}}\] into a simpler form. That is, we have to reduce the exponent of \[{{x}^{8}}\] .
We know that \[{{x}^{8}}\] can be written as the product of \[{{x}^{4}}\] and \[{{x}^{4}}\] .
Now, transforming equation (7), we get
\[\begin{align}
  & ={{\lim }_{x\to 0}}8\times 2\times 2\dfrac{\left\{ \left( {{\sin }^{2}}\dfrac{{{x}^{2}}}{4} \right)\left( {{\sin }^{2}}\dfrac{{{x}^{2}}}{8} \right) \right\}}{\left( {{x}^{4}} \right)\left( {{x}^{4}} \right)} \\
 & ={{\lim }_{x\to 0}}32\dfrac{\left\{ \left( {{\sin }^{2}}\dfrac{{{x}^{2}}}{4} \right)\left( {{\sin }^{2}}\dfrac{{{x}^{2}}}{8} \right) \right\}}{\left( {{x}^{4}} \right)\left( {{x}^{4}} \right)} \\
 & ={{\lim }_{x\to 0}}32\left\{ \dfrac{{{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}^{2}}}{{{\left( {{x}^{2}} \right)}^{2}}}\dfrac{{{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}^{2}}}{{{\left( {{x}^{2}} \right)}^{2}}} \right\} \\
 & ={{\lim }_{x\to 0}}32\left\{ {{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}{4\times \left( \dfrac{{{x}^{2}}}{4} \right)} \right)}^{2}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}{8\times \left( \dfrac{{{x}^{2}}}{8} \right)} \right)}^{2}} \right\} \\
\end{align}\]
\[={{\lim }_{x\to 0}}32\left\{ \dfrac{1}{{{4}^{2}}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}{\left( \dfrac{{{x}^{2}}}{4} \right)} \right)}^{2}}\times \dfrac{1}{{{8}^{2}}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}{\left( \dfrac{{{x}^{2}}}{8} \right)} \right)}^{2}} \right\}\] ………………………………………….(8)
Now, taking \[{{4}^{2}}\] and \[{{8}^{2}}\] out of the bracket and simplifying equation (8), we get
\[={{\lim }_{x\to 0}}32\times \dfrac{1}{{{8}^{2}}\times {{4}^{2}}}\left\{ {{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}{\left( \dfrac{{{x}^{2}}}{4} \right)} \right)}^{2}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}{\left( \dfrac{{{x}^{2}}}{8} \right)} \right)}^{2}} \right\}\]
\[=\dfrac{1}{32}\times {{\lim }_{x\to 0}}\left\{ {{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}{\left( \dfrac{{{x}^{2}}}{4} \right)} \right)}^{2}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}{\left( \dfrac{{{x}^{2}}}{8} \right)} \right)}^{2}} \right\}\]
\[=\dfrac{1}{32}\times {{\lim }_{x\to 0}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{4} \right)}{\left( \dfrac{{{x}^{2}}}{4} \right)} \right)}^{2}}{{\lim }_{x\to 0}}{{\left( \dfrac{\left( \sin \dfrac{{{x}^{2}}}{8} \right)}{\left( \dfrac{{{x}^{2}}}{8} \right)} \right)}^{2}}\] …………………………………………(9)
We know the property that, \[{{\lim }_{x\to 0}}\dfrac{\sin x}{x}=1\] ………………………………………(10)
Now, replacing x by \[\dfrac{{{x}^{2}}}{4}\] in equation (10), we get
\[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{4}}{\dfrac{{{x}^{2}}}{4}}=1\] ……………………………..(11)
Similarly, replacing x by \[\dfrac{{{x}^{2}}}{8}\] in equation (10), we get
\[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{8}}{\dfrac{{{x}^{2}}}{8}}=1\] ……………………………..(12)
From equation (11) and equation (12), we have the value of \[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{4}}{\dfrac{{{x}^{2}}}{4}}\] and \[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{8}}{\dfrac{{{x}^{2}}}{8}}\] .
Now, putting the value of \[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{4}}{\dfrac{{{x}^{2}}}{4}}\] from equation (11) and \[{{\lim }_{x\to 0}}\dfrac{\sin \dfrac{{{x}^{2}}}{8}}{\dfrac{{{x}^{2}}}{8}}\] from equation (12), in equation (9), we get
\[\begin{align}
  & =\dfrac{1}{32}\times 1\times 1 \\
 & =\dfrac{1}{32} \\
\end{align}\]

Therefore, the value of the expression \[{{\lim }_{x\to 0}}\dfrac{8}{{{x}^{8}}}\left( 1-\cos \dfrac{{{x}^{2}}}{2}-\cos \dfrac{{{x}^{2}}}{4}+\cos \dfrac{{{x}^{2}}}{2}\cos \dfrac{{{x}^{2}}}{4} \right)\] is \[\dfrac{1}{32}\].

Note: Whenever there is an expression in which we have to find the value of limit and we have a trigonometric function in the numerator and some function of x in the denominator then always try to transform the expression in the form of \[{{\lim }_{x\to 0}}\dfrac{\sin x}{x}\] .