Question:
The value of
\[
\lim_{x\to0}\frac{\sin5x-5\sin x}{x^3}
\]
is:
The value of
\[
\lim_{x\to0}\frac{\sin5x-5\sin x}{x^3}
\]
is:
Show Hint
Remember the standard expansion
\[
\sin x=x-\frac{x^3}{6}+O(x^5).
\]
It is one of the fastest methods for evaluating difficult trigonometric limits.
Updated On: Jun 10, 2026
- \(-20\)
- \(-10\)
- \(10\)
- \(20\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For limit problems involving trigonometric functions near zero, Maclaurin series expansions are extremely useful.
Recall that
\[
\sin t
=
t-\frac{t^3}{6}+O(t^5).
\]
Using this approximation, complicated trigonometric limits can be transformed into simple algebraic calculations.
Step 1: Expand \(\sin5x\). Using the series, \[ \sin5x = 5x-\frac{(5x)^3}{6}+O(x^5). \] \[ = 5x-\frac{125x^3}{6}+O(x^5). \]
Step 2: Expand \(5\sin x\). \[ 5\sin x = 5\left(x-\frac{x^3}{6}+O(x^5)\right). \] \[ = 5x-\frac{5x^3}{6}+O(x^5). \]
Step 3: Subtract. \[ \sin5x-5\sin x = \left(5x-\frac{125x^3}{6}\right) - \left(5x-\frac{5x^3}{6}\right). \] \[ = -\frac{120x^3}{6}+O(x^5). \] \[ = -20x^3+O(x^5). \]
Step 4: Evaluate the limit. \[ \lim_{x\to0} \frac{-20x^3+O(x^5)}{x^3} = -20. \]
Step 5: Final Conclusion. \[ \boxed{-20} \] Hence the correct answer is \[ \boxed{\text{Option (A)}}. \]
Step 1: Expand \(\sin5x\). Using the series, \[ \sin5x = 5x-\frac{(5x)^3}{6}+O(x^5). \] \[ = 5x-\frac{125x^3}{6}+O(x^5). \]
Step 2: Expand \(5\sin x\). \[ 5\sin x = 5\left(x-\frac{x^3}{6}+O(x^5)\right). \] \[ = 5x-\frac{5x^3}{6}+O(x^5). \]
Step 3: Subtract. \[ \sin5x-5\sin x = \left(5x-\frac{125x^3}{6}\right) - \left(5x-\frac{5x^3}{6}\right). \] \[ = -\frac{120x^3}{6}+O(x^5). \] \[ = -20x^3+O(x^5). \]
Step 4: Evaluate the limit. \[ \lim_{x\to0} \frac{-20x^3+O(x^5)}{x^3} = -20. \]
Step 5: Final Conclusion. \[ \boxed{-20} \] Hence the correct answer is \[ \boxed{\text{Option (A)}}. \]
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