Find the $\lim\limits _{x\to1}\frac{x^3+x-2}{\sin\pi x}$

Question

Find the limit $$\lim_{x\to1}\dfrac{x^3+x-2}{\sin\pi x}$$ I would like to solve the problem without the L'Hôpital's rule. I am also not really familiar with the asymptotic analysis (asymptotic notation).

We can see that $x=1$ is a root of the polynomial in the numerator, and it factors as $(x-1)(x^2+x+2)$. So it's the only real root.

I don't really see how to use the limit of $\dfrac{\sin f(x)}{f(x)}$ when $f(x)$ approaches $0$ (the limit is $1$), as in our problem the argument $\pi x$ of the sine when $x$ approaches $1$ doesn't approach $0$, but $\pi\ne0$. Also how can we get a factor $\pi x$ in the numerator? Is there a general approach for problems of this kind?

Answer 1

\begin{align} \lim_{x\to1}\dfrac{x^3+x-2}{\sin\pi x} &= \lim_{x\to1}\dfrac{(x-1)(x^2+x+2)}{\sin[\pi(x-1)+\pi]} \\ &= \lim_{x\to1}\dfrac{(x-1)\cdot4}{-\sin[\pi(x-1)]} \\ &= -4\lim_{x\to1}\dfrac{x-1}{\sin\pi(x-1)} \\ &= -\frac{4}{\pi}\lim_{x\to1}\dfrac{\pi(x-1)}{\sin\pi(x-1)} \\ &= -\frac{4}{\pi}\lim_{y\to0}\dfrac{y}{\sin y} \\ &= -\frac{4}{\pi}\lim_{y\to0}\dfrac{1}{\frac{\sin y}{y}} \\ &= -\frac{4}{\pi}\dfrac{1}{\lim_{y\to0}\frac{\sin y}{y}} = -\frac{4}{\pi} \end{align}

Answer 2

Using the substitution $y=x-1$ you have

$$\begin{array} \\\lim_{x\to1}\dfrac{x^3+x-2}{\sin\pi x}& \\ =\lim_{x\to1} \dfrac{(x-1)(x^2+x+2)}{\sin\pi x}&\\ =\lim_{y\to0}\dfrac{y(y^2+3y+4)}{\sin\pi (y-1)}&\\ =\lim_{y\to0}\dfrac{y(y^2+3y+4)}{\sin(\pi y-\pi)}&\\ =\lim_{y\to0}\dfrac{\pi}{\pi}\cdot \dfrac{y(y^2+3y+4)}{-\sin\pi y}&\\ =\lim_{y\to0}-\dfrac{y\pi}{\sin\pi y}\cdot \dfrac{(y^2+3y+4)}{\pi}&\\ =-1\cdot \dfrac 4{\pi}=-\dfrac 4{\pi} \end{array}$$