A calculus problem by Bala vidyadharan

The relevant Maclaurin series are

$e^{27x} - 1 = 27x + \dfrac{(27x)^{2}}{2} + \dfrac{(27x)^{3}}{6} + .... = x(27 + O(x^{3}))$ and

$\sin(x) = x - \dfrac{x^{3}}{6} + \dfrac{x^{5}}{120} - .... = x(1 - O(x^{2})).$

The given limit then equals $\lim_{x\to0} \dfrac{27 + O(x^{3})}{1 - O(x^{2})} = \boxed{27}.$

Since plugging in $0$ gives us a $\cfrac{0}{0}$ indeterminate form, we can use L'Hopital's Rule and differentiate the numerator and the denominator: $\begin{aligned} \lim\limits_{x\to 0} \frac{e^{27x}-1}{\sin x} &= \lim\limits_{x\to 0} \frac{\frac{d}{dx}\left(e^{27x}-1\right)}{\frac{d}{dx}\left(\sin x\right)} \\ &= \lim\limits_{x\to 0} \frac{27e^{27x}}{cos x} \\ &= \frac{27e^{27(0)}}{\cos(0)} \\ &= 27e^0 = \green{\boxed{27}} \end{aligned}$