Limits, Continuity & Differentiability Question - None

Question

If $\lim_{x \to 0} \frac{x(1 + a\cos x) - b\sin x}{x^3} = 1$ then

Accepted Answer

Key Idea: Expand numerator using Taylor series and match coefficients of $$x$$ (must be 0) and $$x^3$$ (must equal 1) simultaneously.

Step 1: Taylor expansions:
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots, \quad \sin x = x - \frac{x^3}{6} + \cdots$

Step 2: Expand numerator:
$x\left(1 + a - \frac{ax^2}{2}\right) - b\left(x - \frac{x^3}{6}\right)$ $= x(1+a-b) + x^3\left(-\frac{a}{2} + \frac{b}{6}\right) + \cdots$

Step 3: For limit to be finite (= 1), coefficient of $$x$$ must be zero:
$1 + a - b = 0 \implies b = 1 + a \quad \cdots(1)$

Step 4: Coefficient of $$x^3$$ must equal 1:
$-\frac{a}{2} + \frac{b}{6} = 1 \quad \cdots(2)$

Step 5: Substitute (1) into (2):
$-\frac{a}{2} + \frac{1+a}{6} = 1 \implies -3a + 1 + a = 6 \implies a = -\frac{5}{2}$ Then $b = 1 + (-5/2) = -\dfrac{3}{2}$ .

Answer: (D) $a = -\dfrac{5}{2},\ b = -\dfrac{3}{2}$

Trap: Most students only write one equation (matching

$x^3$

coefficient) and ignore the

$x$

term condition. This gives a wrong system. Both conditions must be applied simultaneously — the limit being finite forces the

$x$

coefficient to vanish first.

Key Concept: Simultaneous coefficient matching using Taylor expansion — two conditions from one limit

Question:

Step-by-Step Solution

Master Limits, Continuity & Differentiability with Mathbee