Evaluate lim_(x→0)(e^4x-1-4x-8x^2)/x^3.

Evaluate $x \to 0 lim \frac{e ^{4 x} - 1 - 4 x - 8 x ^{2}}{x ^{3}}$ .

hard 3 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

\frac{32}{3}

The correct answer is $\frac{32}{3}$ — Hence the required limit is $\frac{32}{3}$ .

Write the Maclaurin series for $e^{4 x}$ : $e^{4 x} = 1 + 4 x + \frac{( 4 x ) ^{2}}{2 !} + \frac{( 4 x ) ^{3}}{3 !} + \dots = 1 + 4 x + 8 x^{2} + \frac{64}{6} x^{3} + \dots$ .
Subtract $1 + 4 x + 8 x^{2}$ from the series: $e^{4 x} - 1 - 4 x - 8 x^{2} = \frac{64}{6} x^{3} + higher powers of x$ .
Divide by $x^{3}$ : $\frac{e ^{4 x} - 1 - 4 x - 8 x ^{2}}{x ^{3}} = \frac{64}{6} + terms containing x$ .
Take the limit as $x \to 0$ ; the higher‑order terms vanish, leaving $\frac{64}{6} = \frac{32}{3}$ .
Hence the required limit is $\frac{32}{3}$ .

Common mistake: Stopping the series after the

x^{2}

term, which eliminates the required

x^{3}

contribution.

Concept tested: Limit using series expansion