Limits, Continuity & Differentiability
Standard Limit — e minus (1+2x)^{1/2x}
nta_pyq_2024_apr
Grade 12
Question:
$\displaystyle\lim_{x\to0}\frac{e-(1+2x)^{\frac{1}{2x}}}{x}$ is equal to
0
$\dfrac{-2}{e}$
$e$
$e-e^2$
Step-by-Step Solution
Key Concept: Write $(1+2x)^{1/(2x)}=e^{\frac{\ln(1+2x)}{2x}}$. Near $x=0$: $\frac{\ln(1+2x)}{2x}=1-x+\frac{4x^2}{3}-\cdots$, so $(1+2x)^{1/(2x)}\approx e^{1-x}\approx e(1-x)$. Then $\frac{e-(1+2x)^{1/(2x)}}{x}\approx\frac{e-e(1-x)}{x}=e$... but from solution it gives $e$.
$\lim_{x\to0}\frac{e-(1+2x)^{1/(2x)}}{x}=e$.
Correct Answer: 3