Evaluate : `lim_(x to 0)(tanx-sinx)/(x^(3))`.

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}, \]

we first notice that substituting \(x = 0\) gives us the indeterminate form \(\frac{0 - 0}{0^3} = \frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Apply L'Hôpital's Rule

We differentiate the numerator and the denominator:

- The derivative of \(\tan x\) is \(\sec^2 x\).
- The derivative of \(\sin x\) is \(\cos x\).
- The derivative of \(x^3\) is \(3x^2\).

Thus, we have:

\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\sec^2 x - \cos x}{3x^2}. \]

### Step 2: Substitute \(x = 0\) Again

Now substituting \(x = 0\) gives us:

\[ \sec^2(0) - \cos(0) = 1 - 1 = 0, \]

and \(3(0^2) = 0\). We again have the form \(\frac{0}{0}\), so we apply L'Hôpital's Rule a second time.

### Step 3: Differentiate Again

Differentiating again:

- The derivative of \(\sec^2 x\) is \(2\sec^2 x \tan x\).
- The derivative of \(\cos x\) is \(-\sin x\).
- The derivative of \(3x^2\) is \(6x\).

Now we have:

\[ \lim_{x \to 0} \frac{2\sec^2 x \tan x + \sin x}{6x}. \]

### Step 4: Substitute \(x = 0\) Again

Substituting \(x = 0\) gives us:

\[ 2\sec^2(0) \tan(0) + \sin(0) = 2(1)(0) + 0 = 0, \]

and \(6(0) = 0\). We still have the form \(\frac{0}{0}\), so we apply L'Hôpital's Rule once more.

### Step 5: Differentiate Again

Differentiating again:

- The derivative of \(2\sec^2 x \tan x\) involves the product rule:
\[ 2(\sec^2 x \sec^2 x + 2\sec^2 x \tan^2 x) = 2\sec^4 x + 4\sec^2 x \tan^2 x. \]
- The derivative of \(\sin x\) is \(\cos x\).
- The derivative of \(6x\) is \(6\).

Now we have:

\[ \lim_{x \to 0} \frac{2\sec^4 x + 4\sec^2 x \tan^2 x + \cos x}{6}. \]

### Step 6: Substitute \(x = 0\) Again

Substituting \(x = 0\) gives:

\[ 2(1) + 4(1)(0) + 1 = 2 + 0 + 1 = 3. \]

Thus, we have:

\[ \lim_{x \to 0} \frac{3}{6} = \frac{1}{2}. \]

### Final Result

Therefore, the limit is:

\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}. \]