Limits Question 11
Question 11 - 2024 (31 Jan Shift 2)
$\lim _{x \rightarrow 0} \frac{a x^{2} e^{x}-b \log _{e}(1+x)+c x e^{-x}}{x^{2} \sin x}=1$ then $16\left(a^{2}+b^{2}+c^{2}\right)$ is equal to
Show Answer
Answer (81)
Solution
$\mathrm{ax}^{2}\left(1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\ldots\right)-\mathrm{b}\left(\mathrm{x}-\frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{x}^{3}}{3}-\ldots \ldots\right)$
$\lim _{x \rightarrow 0} \frac{+c x\left(1-x+\frac{x^{2}}{x !}-\frac{x^{3}}{3 !}+\ldots \ldots\right)}{x^{3} \cdot \frac{\sin x}{x}}$
$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^{2}+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^{3}+\ldots \ldots}{x^{3}}=1$
$c-b=0, \quad \frac{b}{2}-c+a=0$
$\mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2}$
$\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=\frac{9}{16}+\frac{9}{4}+\frac{9}{4}$
$16\left(a^{2}+b^{2}+c^{2}\right)=81$
