Limit of Exponents and Sinusoidal Functions

To find the answer, I will use the following Lemma, which I will leave as an exercise:

Lemma: If $\displaystyle\lim_{x\to 0} f(x) = L$ and $n > 0$ , then $\displaystyle\lim_{x\to 0} (1 + x^n f(x))^{x^{-n}} = e^L$ .

We can rewrite the limit we need to find as follows:

$\displaystyle\lim_{x\to 0} \left[\left(1 + x^2\left(\dfrac{\sin(x)-x}{x^3}\right)\right)^{x^{-2}}\right]^{\dfrac{x^2}{1-\cos(x)}}$

We can apply the Lemma with $n=2$ and $f(x) = \dfrac{\sin(x)-x}{x^3}$ . In this case $\lim_{x\to 0} f(x) = -\dfrac{1}{6}$ (which follows from the Taylor series of $\sin(x)$ ). The Lemma gives us:

$\displaystyle\lim_{x\to 0}\left(1 + x^2\left(\dfrac{\sin(x)-x}{x^3}\right)\right)^{x^{-2}} = e^{-1/6}$

Meanwhile, we have $\displaystyle\lim_{x\to 0}\dfrac{x^2}{1-\cos(x)} = 2$ (this also follows from Taylor series). Hence the limit is equal to $\left(e^{-1/6}\right)^2 = \boxed{e^{-1/3}}$ .

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac {\sin x}x \right)^{\frac 1{1-\cos x}} & \small \color{#3D99F6} \text{A }1^\infty \text{ case (see note).} \\ & = \exp \left(\lim_{x \to 0} \frac 1{1-\cos x}\left(\frac {\sin x}x-1\right)\right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \exp \left(\lim_{x \to 0} \frac {\sin x - x}{x(1-\cos x)} \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{x \to 0} \frac {\cos x - 1}{1-\cos x + x\sin x} \right) & \small \color{#3D99F6} \text{Again, a 0/0 case} \\ & = \exp \left(\lim_{x \to 0} \frac {-\sin x}{\sin x + \sin x + x\cos x} \right) & \small \color{#3D99F6} \text{Again, a 0/0 case} \\ & = \exp \left(\lim_{x \to 0} \frac {-\cos x}{2\cos x + \cos x - x\sin x } \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = e^{-\frac 13} = \frac 1{\sqrt[3]e} \approx 0.716531 \end{aligned}$

$\implies \lfloor 1000L\rfloor = \boxed{716}$

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Note: If $\displaystyle \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = 1^\infty$ , then $\displaystyle \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = \exp \left( h(x) \left(\frac {f(x)}{g(x)} - 1\right) \right)$ . See reference (2nd method) .