Question

$\cos(\tan x) - \cos x \over x$

Answer 1

0

Answer 2

The problem asks to evaluate the limit of the given expression as $x$ approaches $0$: $\lim_{x \to 0} \frac{\cos(\tan x) - \cos x}{x}$

Step 1: Check for indeterminate form. Substitute $x=0$ into the expression: Numerator: $\cos(\tan 0) - \cos 0 = \cos(0) - 1 = 1 - 1 = 0$. Denominator: $0$. Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule. L'Hopital's Rule states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \cos(\tan x) - \cos x$ and $g(x) = x$.

Calculate the derivative of the numerator, $f'(x)$: $f'(x) = \frac{d}{dx}(\cos(\tan x) - \cos x)$ Using the chain rule for $\cos(\tan x)$: $\frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx}$. Here $u = \tan x$, so $\frac{du}{dx} = \sec^2 x$. $f'(x) = -\sin(\tan x) \cdot \sec^2 x - (-\sin x)$ $f'(x) = -\sin(\tan x) \sec^2 x + \sin x$

Calculate the derivative of the denominator, $g'(x)$: $g'(x) = \frac{d}{dx}(x) = 1$

Now, substitute these derivatives into the limit expression: $\lim_{x \to 0} \frac{-\sin(\tan x) \sec^2 x + \sin x}{1}$

Step 3: Evaluate the limit. Substitute $x=0$ into the new expression: As $x \to 0$: $\tan x \to \tan 0 = 0$ $\sin(\tan x) \to \sin(0) = 0$ $\sec x \to \sec 0 = 1$, so $\sec^2 x \to 1^2 = 1$ $\sin x \to \sin 0 = 0$

Therefore, the limit becomes: $-(0) \cdot (1) + 0 = 0 + 0 = 0$

The value of the limit is $0$.