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$f\left(x\right)=\sin \left(x^{-x}\right)e^{\left(x^a\right)}$.

$\lim_{x\to\infty}$ is defined for certain values of $a$.

Find those values.

I noticed that when $a = 0.5$, the function converges.

I also noticed that when $a = -1$, the function converges.

I tried $a = 1.5$ and the graph quickly diverged.

(My logic: Let's say that the function converges for $a \in (-\infty,c)$, there is no reason for $c$ to be some weird irrational logarithm therefore $c=1$, because there are no 'numbers' in my function.)

This is obviously incredibly flawed.

How would I go about proving that this function only has a limit when $a \in (-\infty, 1)$?

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    Try considering $x^{-x}$ instead of $\sin(x^{-x})$: that should yield $\lim_{x\to\infty}e^{x^a-x\ln x}$, and this should be easier to study. – 2017-01-20

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Hints: We have:

(a) $\quad\displaystyle\lim_{x\to +\infty}\sin \left(x^{-x}\right)=0.$

(b) $\quad \displaystyle\lim_{x\to +\infty}x^{\alpha}=\begin{cases}{0}&\text{if}& \alpha <0\\1 & \text{if}& \alpha=0\\+\infty & \text{si}& \alpha>0. \end{cases}\Rightarrow\quad \displaystyle\lim_{x\to +\infty}e^{x^{\alpha}}=\begin{cases}{1}&\text{if}& \alpha <0\\e & \text{if}& \alpha=0\\+\infty & \text{if}& \alpha>0. \end{cases}$

$\qquad\Rightarrow \displaystyle\lim_{x\to +\infty}\sin \left(x^{-x}\right)e^{x^{\alpha}}=\begin{cases}{0}&\text{if}& \alpha <0\\0 & \text{if}& \alpha=0\\0\cdot(+\infty) & \text{if}& \alpha>0. \end{cases}$

(c) For $\alpha >0$, $$L=\lim_{x\to +\infty}\sin \left(x^{-x}\right)e^{x^{\alpha}}=\lim_{x\to +\infty}\dfrac{\sin \left(x^{-x}\right)}{e^{-x^{\alpha}}}=\lim_{x\to +\infty}\dfrac{x^{-x}}{e^{-x^{\alpha}}}.$$

(d) $\quad \log L=\displaystyle\lim_{x\to +\infty}\log\left(\dfrac{x^{-x}}{e^{-x^{\alpha}}}\right).$