Question

If $\underset{x\to 0}{lim}{\left\{2-\cos x\sqrt{\cos 2x}\right\}}^{\frac{x+2}{x^2}}=e^{\alpha }$, then $\alpha =?$

Answer 1

Step 1: Apply limit to the given expression.

$\begin{array}{rcl}\underset{x\to 0}{lim}{\left\{2-\cos x\sqrt{\cos 2x}\right\}}^{\frac{x+2}{x^2}}& =& {\left\{2-\cos 0\sqrt{\cos 0}\right\}}^{\frac{0+2}{0^2}}\\ & =& {\left\{2-1\sqrt{1}\right\}}^{\frac{2}{0}}\\ & =& 1^{\infty }\end{array}$

Step 2: Apply rule to $\underset{x\to 0}{lim}{\left\{2-\cos x\sqrt{\cos 2x}\right\}}^{\frac{x+2}{x^2}}$

We know that if $\underset{x\to 0}{lim}f\left(x\right)=0\mathrm{and}\underset{x\to 0}{lim}g\left(x\right)=\infty$, then $\underset{x\to 0}{lim}{f}^g=e^{\underset{x\to 0}{lim}\left(f-1\right)g}$

So,

$\underset{x\to 0}{lim}{\left\{2-\cos x\sqrt{\cos 2x}\right\}}^{\frac{x+2}{x^2}}=e^{\underset{x\to 0}{lim}\left\{2-\cos x\sqrt{\cos 2x}-1\right\}\left(\frac{x+2}{x^2}\right)}$ …… $\left(1\right)$

Step 3: Solve $\underset{x\to 0}{lim}\left\{2-\cos x\sqrt{\cos 2x}-1\right\}\left(\frac{x+2}{x^2}\right)$

$\begin{array}{rcl}\underset{x\to 0}{lim}\left\{2-\cos x\sqrt{\cos 2x}-1\right\}\left(\frac{x+2}{x^2}\right)& =& \underset{x\to 0}{lim}\frac{\left\{1-\cos x\sqrt{\cos 2x}\right\}\left(x+2\right)}{x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{1-\cos x\sqrt{\cos 2x}\right\}\times {\displaystyle \frac{1+\cos x\sqrt{\cos 2x}}{1+\cos x\sqrt{\cos 2x}}}\times \left(x+2\right)}{x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{1-{\cos }^{2}x\cos 2x\right\}\times \left(x+2\right)}{\left(1+\cos x\sqrt{\cos 2x}\right)x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{1-{\cos }^{2}x\left(1-2{\sin }^{2}x\right)\right\}\times \left(x+2\right)}{\left(1+\cos x\sqrt{\cos 2x}\right)x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{1-{\cos }^{2}x+2{\cos }^{2}x{\sin }^{2}x\right\}\times \left(x+2\right)}{\left(1+\cos x\sqrt{\cos 2x}\right)x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{{\sin }^{2}x+2{\cos }^{2}x{\sin }^{2}x\right\}\times \left(x+2\right)}{\left(1+\cos x\sqrt{\cos 2x}\right)x^2}\\ & =& \underset{x\to 0}{lim}\frac{\left\{{\sin }^{2}x\left(1+2{\cos }^{2}x\right)\right\}\times \left(x+2\right)}{\left(1+\cos x\sqrt{\cos 2x}\right)x^2}\\ & =& \underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}\times \frac{\left\{\left(1+2{\cos }^{2}x\right)\right\}}{\left(1+\cos x\sqrt{\cos 2x}\right)}\times \left(x+2\right)\\ & =& \underset{x\to 0}{lim}1\times \frac{\left\{\left(1+2{\cos }^{2}x\right)\right\}}{\left(1+\cos x\sqrt{\cos 2x}\right)}\times \left(x+2\right);\left(by\underset{x\to 0}{lim}\frac{\sin x}{x}=1\right)\end{array}$

By applying limit we get

$$\begin{gathered} \frac{1\times \left\{\left(1+2{\cos }^{2}0\right)\right\}\times \left(0+2\right)}{\left(1+\cos 0\sqrt{\cos 0}\right)} \\ =\frac{1\times 3\times 2}{1+1} \\ =3 \end{gathered}$$

Step 4: Find the value of $\alpha$.

By substituting the value of $\underset{x\to 0}{lim}\left\{2-\cos x\sqrt{\cos 2x}-1\right\}\left(\frac{x+2}{x^2}\right)$ in $\left(1\right)$, we get

$\underset{x\to 0}{lim}{\left\{2-\cos x\sqrt{\cos 2x}\right\}}^{\frac{x+2}{x^2}}=e^3$

That means $e^{\alpha }=e^3$.

Therefore, $\alpha =3$