Limits, Continuity & Differentiability
Taylor Series Expansion — Finding Coefficients from Limit
nta_pyq_2024_jan
Grade 12
Question:
If $\displaystyle\lim_{x\to0}\dfrac{3+\alpha\sin x+\beta\cos x+\log_e(1-x)}{3\tan^2x}=\dfrac{1}{3}$, then $2\alpha-\beta$ is equal to:
Step-by-Step Solution
Key Concept: Expand $\sin x$, $\cos x$, $\log_e(1-x)$ using Taylor series. For the limit to exist (finite and non-zero) as $x\to0$ with denominator $\sim3x^2$: constant term must vanish: $3+\beta+0=0$, linear term must vanish: $\alpha-1=0$. Then the coefficient of $x^2$ over $3$ equals $1/3$.
$\beta=-3$, $\alpha=1$. $2\alpha-\beta=2-(-3)=5$.
Correct Answer: 3