Question

If the value of $\lim_{x\to0}(2-\cos x\sqrt{\cos 2x})^{(\frac{x+2}{x^2})}$ is equal to $e^a$, then a is equal to

Answer 1

3

Answer 2

The given limit is of the form $\lim_{x\to0}(2-\cos x\sqrt{\cos 2x})^{(\frac{x+2}{x^2})}$.

Let $f(x) = 2-\cos x\sqrt{\cos 2x}$ and $g(x) = \frac{x+2}{x^2}$.

As $x \to 0$: $f(x) \to 2 - \cos(0)\sqrt{\cos(0)} = 2 - 1 \cdot \sqrt{1} = 2 - 1 = 1$. $g(x) \to \frac{0+2}{0^2} = \frac{2}{0} \to \infty$.

This is an indeterminate form of type $1^\infty$. For limits of the form $\lim_{x\to a} [f(x)]^{g(x)}$ which are $1^\infty$, we use the formula: $L = e^{\lim_{x\to a} g(x)[f(x)-1]}$ In this case, $f(x)-1 = (2-\cos x\sqrt{\cos 2x}) - 1 = 1-\cos x\sqrt{\cos 2x}$. So, the exponent is $A = \lim_{x\to0} \frac{x+2}{x^2} (1-\cos x\sqrt{\cos 2x})$. We can write this as: $A = \left(\lim_{x\to0} (x+2)\right) \cdot \left(\lim_{x\to0} \frac{1-\cos x\sqrt{\cos 2x}}{x^2}\right)$ The first limit is $0+2 = 2$. Let's evaluate the second limit, $H = \lim_{x\to0} \frac{1-\cos x\sqrt{\cos 2x}}{x^2}$. As $x \to 0$, the numerator $1-\cos(0)\sqrt{\cos(0)} = 1-1=0$, and the denominator $0^2=0$. So, it is a $\frac{0}{0}$ indeterminate form, and we can apply L'Hopital's Rule.

Let $N(x) = 1-\cos x\sqrt{\cos 2x}$. Let $D(x) = x^2$.

Differentiating $N(x)$: $N'(x) = - \left( (-\sin x)\sqrt{\cos 2x} + \cos x \cdot \frac{1}{2\sqrt{\cos 2x}} \cdot (-\sin 2x \cdot 2) \right)$ $N'(x) = \sin x\sqrt{\cos 2x} + \frac{\cos x \sin 2x}{\sqrt{\cos 2x}}$ Using $\sin 2x = 2\sin x \cos x$: $N'(x) = \sin x\sqrt{\cos 2x} + \frac{\cos x (2\sin x \cos x)}{\sqrt{\cos 2x}}$ $N'(x) = \sin x \left( \sqrt{\cos 2x} + \frac{2\cos^2 x}{\sqrt{\cos 2x}} \right)$ $N'(x) = \sin x \left( \frac{\cos 2x + 2\cos^2 x}{\sqrt{\cos 2x}} \right)$ Using the identity $\cos 2x = 2\cos^2 x - 1$, we have $\cos 2x + 2\cos^2 x = (2\cos^2 x - 1) + 2\cos^2 x = 4\cos^2 x - 1$. So, $N'(x) = \sin x \frac{4\cos^2 x - 1}{\sqrt{\cos 2x}}$.

Differentiating $D(x)$: $D'(x) = 2x$.

Now, apply L'Hopital's Rule: $H = \lim_{x\to0} \frac{N'(x)}{D'(x)} = \lim_{x\to0} \frac{\sin x \frac{4\cos^2 x - 1}{\sqrt{\cos 2x}}}{2x}$ $H = \lim_{x\to0} \left(\frac{\sin x}{x}\right) \cdot \frac{1}{2} \cdot \left(\frac{4\cos^2 x - 1}{\sqrt{\cos 2x}}\right)$ We know that $\lim_{x\to0} \frac{\sin x}{x} = 1$. Substitute $x=0$ into the remaining terms: $H = 1 \cdot \frac{1}{2} \cdot \frac{4\cos^2(0) - 1}{\sqrt{\cos(0)}}$ $H = \frac{1}{2} \cdot \frac{4(1)^2 - 1}{\sqrt{1}} = \frac{1}{2} \cdot \frac{4-1}{1} = \frac{1}{2} \cdot 3 = \frac{3}{2}$ Now, substitute this value of $H$ back into the expression for $A$: $A = 2 \cdot H = 2 \cdot \frac{3}{2} = 3$ Therefore, the original limit $L = e^A = e^3$. Given that $L = e^a$, we have $e^a = e^3$. Thus, $a=3$.