Limits, Continuity & Differentiability
Taylor Expansion — Three-Parameter Limit
nta_pyq_2024_jan
Grade 12
If $\displaystyle\lim_{x\to0}\dfrac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1$, then $16(a^2+b^2+c^2)$ is equal to
Step-by-Step Solution
Key Concept: Expand $e^x=1+x+x^2/2!+\ldots$, $\log_e(1+x)=x-x^2/2+x^3/3-\ldots$, $e^{-x}=1-x+x^2/2!-\ldots$ Collect coefficients of $x^0$, $x^1$, $x^2$ in numerator. For the limit to be finite, coefficients of $x^0$ and $x^1$ must vanish; coefficient of $x^2$ divided by $1$ (since $x^2\sin x\sim x^3$)... denominator $\sim x^3$, so numerator coefficient of $x^3$ gives the limit.
Taylor expansion gives $c=b$, $b/2-c+a=0$, $a-b/3+c/2=1$. Solving: $a=3/4$, $b=c=3/2$. $16(9/16+9/4+9/4)=16\times81/16=81$.
Correct Answer: 81