A question about computing limit

Question

I came across this exercise

$f(x,y)= \lim_{y\to\infty}{{1-y\sin{\pi x\over y}}\over \arctan x}$

The result I get is ${1-\pi x \over \arctan x}$, which depends on the value of $x$.

However, the question I have is that whatever $x$ is, since it's in the $\sin()$, which is a bounded function, shouldn't lay any effect on the limit. For example: $\lim_{x\to\infty}{1 \over x}\sin ax = 0, a\in(-\infty,+\infty)$, whose limit doesn't depend on $a$.

Is there any intuitive understanding for this ?

Answer 1

You appear to be under the delusion that the correct way to apply the change of variable $y=\frac1x$ to $\lim_{y\to \infty} y\sin\frac{a}{y}$ is $$\lim_{x\to\infty}\frac1x\sin(ax)$$ while it should rather be $$\lim_{x\to 0}\frac1x\sin(ax)$$ (if we wanted to be pedantic, it should actually be $x\to 0^+$, but whatever).

Answer 2

Hint

Let $\frac 1y=t$ Then as $y\to \infty$ hence $t\to 0$ Hence $$\lim_{y\to \infty} y\sin \frac {\pi x}{y}=\lim_{y\to \infty} \frac {\sin \frac {\pi x}{y}}{\frac 1y}=\lim_{t=0} \frac {\sin \pi xt}{t}=\pi x$$

Answer 3

You are finding the limit as y approaches infinity at a fixed point x. Thus $$f(x,y)= \lim_{y\to\infty}{{1-y\sin{\pi x\over y}}\over arctanx }={1-\pi x \over acrtanx}$$ is expected.

Of course the limit should depend on $x$ and it varies with x, but that does not mean that there is something wrong with the limit.