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

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

& ={{\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{aligned}$$ $$={{\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{aligned} & =\dfrac{1}{32}\times 1\times 1 \\\ & =\dfrac{1}{32} \\\ \end{aligned}$$ **Therefore, the value of the expression $${{\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}$$ .