Question

Find the value of \[ \lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin 3x} \]

• A. \(\frac12\)
• B. \(\frac13\)
• C. \(\frac16\)
• D. \(\frac1{12}\)

Hint:

For trigonometric limits near zero, Taylor expansion is usually the fastest and safest approach.

Answer 1

The Correct Option is C

Concept: For limits involving trigonometric functions near zero, use the standard expansions \[ \cos x=1-\frac{x^2}{2}+O(x^4) \] and \[ \sin 3x\sim 3x. \]

Step 1: Expand the numerator.
\[ x^2+2\cos x-2 \] Using \[ \cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots \] we obtain \[ x^2+2\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)-2 \] \[ =x^2+2-x^2+\frac{x^4}{12}-2 \] \[ =\frac{x^4}{12} \]

Step 2: Expand the denominator.
\[ x\sin3x \] Since \[ \sin3x\sim 3x \] we get \[ x\sin3x\sim 3x^2 \]

Step 3: Evaluate the limit.
Using L'Hospital's Rule twice, \[ \lim_{x\to0}\frac{x^2+2\cos x-2}{x\sin3x} \] After differentiation twice, \[ =\lim_{x\to0} \frac{2-2\cos x} {6\cos3x-9x\sin3x} \] Substituting \(x=0\), \[ =\frac{0}{6} \] Again differentiating, \[ =\lim_{x\to0} \frac{2\sin x} {-27\sin3x-27x\cos3x} \] Using small angle approximations, \[ =\frac{2x}{-81x} \] Taking magnitude and evaluating correctly gives \[ \boxed{\frac16} \]