Limit of $\sin(\pi x^a)/\sin(\pi x^b)$ as $x\rightarrow 1$ without l'Hôpital's rule?

Question

How can I solve this limit without using l'Hopital's rule or any other method, which uses derivation, please?

$$\lim_{x\rightarrow 1} \frac{\sin(\pi x^a)}{\sin(\pi x^b)}$$

$$a,b\in \mathbb{R}, b \neq 0$$

Answer 1

Note that $$\lim_{x \to 1}\frac{\sin(\pi x^a)}{\sin(\pi x^b)}=\lim_{x \to 1}\frac{\sin(\pi(x^a-1))}{\pi(x^a-1)}\cdot\frac{\pi(x^b-1)}{\sin(\pi(x^b-1))}\frac{x^a-1}{x^b-1}. \tag 1$$ Using $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta}=1$, $(1)$ simplifies to $$\lim_{x \to 1}\frac{\sin(\pi x^a)}{\sin(\pi x^b)} = \lim_{x \to 1}\frac{x^a-1}{x^b-1}=\frac{a}{b},$$ where the last limit comes from this answer.

Answer 2

Is it in your best interest to use the exponential form of the $\sin$ function?

For example you can express

$$ \lim_{x \rightarrow 1} \frac{\sin(\pi x^a)}{\sin(\pi x^b)} = \lim_{x \rightarrow 1} \frac{e^{-i \pi x^a} - e^{i \pi x^a}}{e^{-i \pi x^b} - e^{i π x^b}}.$$

Answer 3

Let $\\ x-1=t\\ x=t+1$ so that $$ \\ \\ \lim _{ x\rightarrow 1 } \frac { \sin (\pi x^{ a }) }{ \sin (\pi x^{ b }) } =\lim _{ t\rightarrow 0 } \frac { \sin (\pi \left( t+1 \right) ^{ a }) }{ \sin (\pi \left( t+1 \right) ^{ b }) } =\lim _{ t\rightarrow 0 } \frac { \sin (\pi \left( t+1 \right) ^{ a }) }{ \sin (\pi \left( t+1 \right) ^{ b }) } =\\ =\lim _{ t\rightarrow 0 } \frac { \sin \left( \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+1 \right) \right) }{ \sin \left( \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+1 \right) \right) } =\lim _{ t\rightarrow 0 } \frac { \sin \left( \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+at \right) \right) }{ \sin \left( \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+bt \right) \right) } =\\ =\lim _{ t\rightarrow 0 } \frac { \sin \left( \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+at \right) \right) }{ \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+at \right) } \frac { \left( \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+bt \right) \right) }{ \sin \left( \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+bt \right) \right) } \frac { \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+at \right) }{ \left( \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+bt \right) \right) } =\\ =\lim _{ t\rightarrow 0 } \frac { \pi \left( { t }^{ a }+a{ t }^{ a-1 }+...+at \right) }{ \pi \left( { t }^{ b }+b{ t }^{ b-1 }+...+bt \right) } =\lim _{ t\rightarrow 0 } \frac { t\left( { t }^{ a-1 }+a{ t }^{ a-2 }+...+a \right) }{ t\left( { t }^{ b-1 }+b{ t }^{ b-2 }+...+b \right) } =\frac { a }{ b } \\ $$

It is not a general solution, we can apply it,in case $a$ and $b$ are positive integers(because here have been used "Binomial theorem" )

Answer 4

Here I solve for the left limit and $a>1$. For general setting should be followed in the similar fashion. By $x=1-y$ for small $y>0$, so I concentrate on the limit $\lim_{y\rightarrow 0^{+}}\dfrac{\sin(\pi(1-y)^{a})}{\pi ay}$ and I claim that the value is $1$.

I make use two inequalities. First, for small $y>0$, $(1-y)^{a}\geq 1-ay$. Second, for small $y>0$, $(1-y)^{a}\leq 1-ay+\dfrac{a(a-1)}{2}y^{2}$. Locally for $x=\pi$, $\sin x$ is decreasing, so \begin{align*} \frac{\sin\left(\pi\left(1-ay+\dfrac{a(a-1)}{2}y^{2}\right)\right)}{\pi ay}\leq\frac{\sin(\pi(1-y)^{a})}{\pi ay}\leq\frac{\sin(\pi(1-ay))}{\pi ay}, \end{align*} take into account that $\sin(\pi(1-ay))=\sin(\pi ay)$ and $\sin\left(\pi\left(1-ay+\dfrac{a(a-1)}{2}y^{2}\right)\right)=\sin\left(\pi ay-\pi\dfrac{a(a-1)}{2}y^{2}\right)$, one can squeeze the limits into the value $1$.

For the question, one can see that $\dfrac{\sin(\pi(1-y)^{a})}{\sin(\pi(1-y)^{b})}=\dfrac{a}{b}\dfrac{\sin(\pi(1-y)^{a})}{\pi ay}\dfrac{\pi by}{\sin(\pi(1-y)^{b})}$.