Find the value of `lim_(x rarr 0) ( sin x - 2 sin 3x + sin5x)/(x) `

Options

A. `-1`
B. 0
C. 1
D. 3

Answer

To find the limit

\[ \lim_{x \to 0} \frac{\sin x - 2 \sin 3x + \sin 5x}{x} \]

we first observe that substituting \(x = 0\) directly into the expression yields an indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator separately.

**Step 1: Differentiate the numerator and denominator.**

The numerator is \( \sin x - 2 \sin 3x + \sin 5x \). We differentiate each term:

- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( -2 \sin 3x \) is \( -2 \cdot 3 \cos 3x = -6 \cos 3x \).
- The derivative of \( \sin 5x \) is \( 5 \cos 5x \).

Thus, the derivative of the numerator is:

\[ \cos x - 6 \cos 3x + 5 \cos 5x \]

The derivative of the denominator \(x\) is simply \(1\).

**Step 2: Rewrite the limit using the derivatives.**

Now we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\cos x - 6 \cos 3x + 5 \cos 5x}{1} \]

**Step 3: Substitute \(x = 0\) into the new limit.**

Now we substitute \(x = 0\):

- \(\cos 0 = 1\)
- \(\cos 3 \cdot 0 = \cos 0 = 1\)
- \(\cos 5 \cdot 0 = \cos 0 = 1\)

So we have:

\[ 1 - 6 \cdot 1 + 5 \cdot 1 = 1 - 6 + 5 = 0 \]

**Step 4: Since we still have an indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule again.**

We differentiate the numerator again:

- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( -6 \cos 3x \) is \( 18 \sin 3x \).
- The derivative of \( 5 \cos 5x \) is \( -25 \sin 5x \).

Thus, the second derivative of the numerator is:

\[ -\sin x + 18 \sin 3x - 25 \sin 5x \]

The derivative of the denominator is still \(1\).

**Step 5: Rewrite the limit again.**

Now we rewrite the limit as:

\[ \lim_{x \to 0} (-\sin x + 18 \sin 3x - 25 \sin 5x) \]

**Step 6: Substitute \(x = 0\) into the new limit.**

Substituting \(x = 0\):

- \(-\sin 0 = 0\)
- \(18 \sin 3 \cdot 0 = 0\)
- \(-25 \sin 5 \cdot 0 = 0\)

So we have:

\[ 0 + 0 - 0 = 0 \]

**Step 7: Since we still have an indeterminate form \(0\), we apply L'Hôpital's Rule a third time.**

Now we differentiate the numerator again:

- The derivative of \(-\sin x\) is \(-\cos x\).
- The derivative of \(18 \sin 3x\) is \(54 \cos 3x\).
- The derivative of \(-25 \sin 5x\) is \(-125 \cos 5x\).

Thus, the third derivative of the numerator is:

\[ -\cos x + 54 \cos 3x - 125 \cos 5x \]

The denominator remains \(1\).

**Step 8: Rewrite the limit again.**

Now we rewrite the limit as:

\[ \lim_{x \to 0} (-\cos x + 54 \cos 3x - 125 \cos 5x) \]

**Step 9: Substitute \(x = 0\) into the new limit.**

Substituting \(x = 0\):

- \(-\cos 0 = -1\)
- \(54 \cos 3 \cdot 0 = 54 \cdot 1 = 54\)
- \(-125 \cos 5 \cdot 0 = -125 \cdot 1 = -125\)

So we have:

\[ -1 + 54 - 125 = -1 + 54 - 125 = -72 \]

Thus, the value of the limit is:

\[ \boxed{-72} \]