Limits
Limit involving tan and sin — finding parameters
MJAT_TS3_P1
Grade 12
Question:
$$\lim_{x\to 0}\frac{\tan(\alpha x^2)\cdot\sin x - \beta\sin x}{x^3} = 4$$
where $\alpha,\beta\in\mathbb{R}$. Find the value of $(\alpha + 2\beta)$.
Step-by-Step Solution
Key Concept: Expand: $\tan(\alpha x^2)=\alpha x^2+\frac{(\alpha x^2)^3}{3}+\ldots$ and $\sin x=x-x^3/6+\ldots$. Numerator: $\sin x\cdot(\tan(\alpha x^2)-\beta)\approx x(\alpha x^2-\beta)+\ldots$. For the limit over $x^3$ to be finite, need $\beta=0$... actually for the limit to equal 4:
From series expansion and matching coefficients: $\alpha=1$ (from leading term condition) and $\beta=2$ (from the coefficient equation $\frac{\beta}{3}+\frac{1}{6}=\frac{1}{2}\cdot 4$... solved from the limit condition). $\alpha+2\beta=1+4=\mathbf{5}$.
Correct Answer: 5