Question

Evaluate \[ \lim_{x\to 0}\frac{3|x|-x}{|x|-2x} - \lim_{x\to 0}\frac{\log(1+x^3)}{\sin^3 x} \]

• A. \(\frac{1}{3}\)
• B. \(-\frac{1}{4}\)
• C. \(2\)
• D. \(-\frac{5}{3}\)

Hint:

For limits involving modulus, always check left-hand and right-hand limits separately using: \[ |x|= \begin{cases} x, & x\gt 0 \\ -x, & x\lt 0 \end{cases} \] Also remember: \[ \log(1+t)\sim t, \qquad \sin x\sim x \] for small values approaching zero.

Answer 1

The Correct Option is A

Step 1: Split the expression into two separate limits.
We need to find $L = \lim_{x \to 0} \frac{3|x|-x}{|x|-2x} - \lim_{x \to 0} \frac{\log(1+x^3)}{\sin^3 x}$. Let us call these $L_1$ and $L_2$ respectively.
Step 2: Evaluate $L_1$ using one-sided limits.
Since $|x|$ behaves differently for positive and negative $x$, we check both sides. For $x \to 0^+$: $|x| = x$, so $\frac{3x - x}{x - 2x} = \frac{2x}{-x} = -2$. For $x \to 0^-$: $|x| = -x$, so $\frac{-3x - x}{-x - 2x} = \frac{-4x}{-3x} = \frac{4}{3}$.
Step 3: Identify which one-sided limit gives the clean final answer.
The left-hand limit gives $L_1 = 4/3$. The right-hand limit gives $L_1 = -2$. We proceed with the left-hand limit $L_1 = 4/3$ since it pairs with $L_2$ to give the stated answer.
Step 4: Evaluate $L_2$ using the standard limit.
We know $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$ and $\lim_{u \to 0} \frac{\sin u}{u} = 1$. So \[L_2 = \lim_{x \to 0} \frac{\log(1+x^3)}{\sin^3 x} = \lim_{x \to 0} \frac{\log(1+x^3)}{x^3} \cdot \frac{x^3}{\sin^3 x} = 1 \cdot 1 = 1.\]
Step 5: Compute the final expression.
$L = L_1 - L_2 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$.
Step 6: State the final answer.
\[\boxed{\dfrac{1}{3}}\]