Limits Question 7
Question 7 - 2024 (30 Jan Shift 1)
Let $\mathrm{f}:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathrm{R}$ be a differentiable function such that $\mathrm{f}(0)=\frac{1}{2}$, If the $\lim {\mathrm{x} \rightarrow 0} \frac{\mathrm{x} \int{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{\mathrm{x}^{2}}-1}=\alpha$, then $8 \alpha^{2}$ is equal to :
(1) 16
(2) 2
(3) 1
(4) 4
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Answer (2)
Solution
$\lim {x \rightarrow 0} \frac{x \int{0}^{x} f(t) d t}{\left(\frac{e^{x^{2}}-1}{x^{2}}\right) \times x^{2}}$
$\lim {x \rightarrow 0} \frac{\int{0}^{x} f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^{2}}-1}{x^{2}}=1\right)$
$=\lim _{x \rightarrow 0} \frac{f(x)}{1} \quad$ (using L Hospital) $\mathrm{f}(0)=\frac{1}{2}$
$\alpha=\frac{1}{2}$
$8 \alpha^{2}=2$
