Plz solve this lovely problem

Let $a,b$ be real constants satisfying $\displaystyle \lim_{x\to0} \left( \dfrac1{bx - e \sin x} \int_0^x \dfrac{t^2}{\sqrt{a + 2t^5}} \, dt \right ) = \dfrac1\pi.$ Find the value of $a$ .

#Calculus

Easy Math Editor

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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Hint: Apply Fundamental theorem of calculus and L-hopital's rule.

Hint 2: What happens to the limit when $b \ne e$ ?

Hint 3: $1 - \cos x = 2\sin^2 (x/2)$ . Divide numerator and denominator by $x^2$ .

Hint 4: Apply $\displaystyle \lim_{x\to0} \dfrac{\sin x}x = 1$ .

Pi Han Goh - 3 years, 3 months ago

It’s not lovely

Annie Li - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$