What happens to the following limit when $b\in [0, 1)$ and $b > 1$? Problem:
Find all $a, b$ which make the following statement true: $\lim_{x\to 0}\frac{\exp{\left(\sin ax\right)} - \cos x}{x^b} = \frac{1}{2}$
Attempted solution:
Firstly, let's notice that if $b < 0$ then the limit is $0$ no matter what the value of $a$ is.
Secondly, let's rewrite the fraction using the Maclaurin Series for $\exp$ and $\cos$:
$$
\frac{1 + \sin(ax) + \frac{1}{2!}(\sin^2 ax) + \frac{1}{3!}\sin^3(ax) + \dots+ (-1)(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots)}{x^b} = \\
\frac{\sin(ax) + \frac{1}{2!}(\sin^2(ax) + x^2) + \frac{1}{3!}\sin^3(ax) + \frac{1}{4!}(\sin^4(ax) - x^4) + \dots}{x^b}
$$
From this representation it is clear that if $b = 1$ and $a = 0.5$:
$$
\begin{split}
&\lim_{x\to 0}\frac{\sin(0.5 x) + \frac{1}{2!}(\sin^2(0.5 x) + x^2) + \frac{1}{3!}\sin^3(0.5x) + \frac{1}{4!}(\sin^4(0.5x) - x^4) + \dots}{x} &= \\
&\lim_{x\to 0}\frac{\sin(0.5 x)}{x} + 0 &= \frac{1}{2}
\end{split}
$$
At this point let's remember that $\lim_{x\to 0}\sin x = \lim_{x\to 0} x$ so the fraction may be rewritten like so:
$$
\lim_{x\to 0}\frac{ax + \frac{1}{2!}((ax)^2 + x^2) + \frac{1}{3!}(ax)^3 + \frac{1}{4!}((ax)^4 - x^4) + \dots}{x^b}
$$
which makes this limit $0$ if $b\in[0, 1)$ and $\infty$ if $b > 1$.
 A: Hint: By using limiting arguments about the behavior of functions $e^x$, $\sin x$, and $\cos x$ for small $x$ (e.g. using Taylor series), you should be able to show that your limit is equal to $\lim_{x \to 0} \frac{ax}{x^b}$
A: If $b=0$ then the limit is $0$ independently of $a$.
Note that no matter how $a$ and $b>0$ are, this limit is of the type $\frac 0 0$, therefore let us use l'Hospital's theorem on it, replacing the numerator and the denominator by their derivatives, respectively: $\lim \limits _{x \to 0} \frac {a \Bbb e ^{\sin ax} \cos ax + \sin x} {b x^{b-1}} = \lim \limits _{x \to 0} \frac {\cos ax} b \frac {a \Bbb e ^{\sin ax} + \tan x} {x^{b-1}} = \frac 1 b \lim \limits _{x \to 0} \frac {a \Bbb e ^{\sin ax} + \tan x} {x^{b-1}}$. Note that the numerator tends to $a$.
If $b<1$ then $x$ must be positive (what would $(-1) ^{0.5}$ mean?), so the denominator tends to $\infty$, so the whole limit is $0$. If $b=1$ then the denominator is constant $1$ and the limit is $\frac a b$. Finally, if $b>1$ you must have again $x>0$ and the limit is $\frac 1 b \frac 1 {0^+} = \infty$.
Therefore, the only possibility is $a=\frac 1 2, b=1$.
