Question

$\qquad L = \lim_{x \to 0} \frac{2(\csc x - n \cot 2x)}{x^3(\cot x \csc x - \sec x)}$ = some integer and $n \in I$, then the value of $n + L$ is ____.

Answer 1

5

Answer 2

To evaluate the limit $L = \lim_{x \to 0} \frac{2(\csc x - n \cot 2x)}{x^3(\cot x \csc x - \sec x)}$, we first simplify the expression by converting trigonometric functions into $\sin x$ and $\cos x$.

Step 1: Simplify the numerator

Numerator $N(x) = 2(\csc x - n \cot 2x)$

$N(x) = 2\left(\frac{1}{\sin x} - n \frac{\cos 2x}{\sin 2x}\right)$

Using $\sin 2x = 2 \sin x \cos x$:

$N(x) = 2\left(\frac{1}{\sin x} - n \frac{\cos 2x}{2 \sin x \cos x}\right)$

$N(x) = \frac{2}{\sin x}\left(1 - \frac{n \cos 2x}{2 \cos x}\right)$

$N(x) = \frac{2}{\sin x}\left(\frac{2 \cos x - n \cos 2x}{2 \cos x}\right)$

$N(x) = \frac{2 \cos x - n \cos 2x}{\sin x \cos x}$

Step 2: Simplify the denominator

Denominator $D(x) = x^3(\cot x \csc x - \sec x)$

$D(x) = x^3\left(\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} - \frac{1}{\cos x}\right)$

$D(x) = x^3\left(\frac{\cos x}{\sin^2 x} - \frac{1}{\cos x}\right)$

$D(x) = x^3\left(\frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos x}\right)$

Using $\cos^2 x - \sin^2 x = \cos 2x$:

$D(x) = x^3\left(\frac{\cos 2x}{\sin^2 x \cos x}\right)$

Step 3: Substitute simplified expressions into the limit

$L = \lim_{x \to 0} \frac{\frac{2 \cos x - n \cos 2x}{\sin x \cos x}}{x^3\left(\frac{\cos 2x}{\sin^2 x \cos x}\right)}$

$L = \lim_{x \to 0} \frac{2 \cos x - n \cos 2x}{\sin x \cos x} \cdot \frac{\sin^2 x \cos x}{x^3 \cos 2x}$

$L = \lim_{x \to 0} \frac{(2 \cos x - n \cos 2x) \sin x}{x^3 \cos 2x}$

We can rewrite this as:

$L = \lim_{x \to 0} \frac{2 \cos x - n \cos 2x}{x^2} \cdot \frac{\sin x}{x} \cdot \frac{1}{\cos 2x}$

Step 4: Evaluate the limit using standard limit forms

As $x \to 0$, we know:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

$\lim_{x \to 0} \cos 2x = \cos 0 = 1$

So, the limit simplifies to:

$L = \lim_{x \to 0} \frac{2 \cos x - n \cos 2x}{x^2} \cdot 1 \cdot 1$

$L = \lim_{x \to 0} \frac{2 \cos x - n \cos 2x}{x^2}$

Step 5: Determine the value of n

For the limit to be finite (as $L$ is an integer), the numerator must approach 0 as $x \to 0$ (since the denominator approaches 0).

Let $f(x) = 2 \cos x - n \cos 2x$.

$f(0) = 2 \cos 0 - n \cos 0 = 2(1) - n(1) = 2 - n$.

For the limit to be of the form $\frac{0}{0}$, we must have $2 - n = 0$, which implies $n = 2$.

Step 6: Calculate the value of L using n=2

Substitute $n=2$ into the limit expression:

$L = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2x}{x^2}$

This is of the form $\frac{0}{0}$, so we can apply L'Hopital's Rule.

Differentiate the numerator and denominator with respect to $x$:

$\frac{d}{dx}(2 \cos x - 2 \cos 2x) = -2 \sin x - 2(-\sin 2x)(2) = -2 \sin x + 4 \sin 2x$

$\frac{d}{dx}(x^2) = 2x$

So, $L = \lim_{x \to 0} \frac{-2 \sin x + 4 \sin 2x}{2x}$

$L = \lim_{x \to 0} \left(\frac{-2 \sin x}{2x} + \frac{4 \sin 2x}{2x}\right)$

$L = \lim_{x \to 0} \left(-\frac{\sin x}{x} + 2 \frac{\sin 2x}{x}\right)$

To use the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$, we adjust the second term:

$L = \lim_{x \to 0} \left(-\frac{\sin x}{x} + 4 \frac{\sin 2x}{2x}\right)$

$L = -(1) + 4(1)$

$L = -1 + 4 = 3$.

Alternatively, using Taylor series expansion:

$\cos x = 1 - \frac{x^2}{2!} + O(x^4)$

$\cos 2x = 1 - \frac{(2x)^2}{2!} + O(x^4) = 1 - 2x^2 + O(x^4)$

Numerator $= 2 \cos x - 2 \cos 2x = 2\left(1 - \frac{x^2}{2}\right) - 2(1 - 2x^2) + O(x^4)$

$= (2 - x^2) - (2 - 4x^2) + O(x^4)$

$= 3x^2 + O(x^4)$

$L = \lim_{x \to 0} \frac{3x^2 + O(x^4)}{x^2} = \lim_{x \to 0} (3 + O(x^2)) = 3$.

Step 7: Calculate n+L

We found $n=2$ and $L=3$.

$n+L = 2+3 = 5$.

The final answer is $\boxed{5}$.