Limits 10

Consider $f(x)=(\sin x-\cos x)$ and $g(x)=\dfrac{1}{x^3}$ . If you consider the given limit as $L$ and the left and right hand limits of $L$ as $L_{left}$ and $L_{right}$ , then, we can see that,

$L=\lim_{x\to 0} f(x)g(x)~,~L_{left}=\lim_{x\to 0^-} f(x)g(x)~,~L_{right}=\lim_{x\to 0^+} f(x)g(x)$

Now, note that neither $f(x)$ nor $g(x)$ approach $0$ as $x\to 0^+,0^-$ , so we can use the product rule to separate the limits as follows:

$L_{left}=\left(\lim_{x\to 0^-}f(x)\right)\left(\lim_{x\to 0^-}g(x)\right)~,~L_{right}=\left(\lim_{x\to 0^+}f(x)\right)\left(\lim_{x\to 0^+}g(x)\right)$

Also, we have, $\displaystyle\lim_{x\to 0} f(x)=\lim_{x\to 0} (\sin x-\cos x)=(0-1)=(-1)$

Now, we have a decently well known result that,

$\lim_{x\to 0^{\large \pm}}\frac{1}{x^m}=\begin{cases}\pm\infty~,~\textrm{when }m\textrm{ is odd}\\ \infty~,~\textrm{when }m\textrm{ is even}\end{cases}~,~m\in\mathbb{Z^+}$

In our case, $g(x)$ resembles this form with $m=3$ which is odd and as such, we have,

$L_{left}=(-1)\cdot (-\infty)=(+\infty)~,~L_{right}=(-1)\cdot (+\infty)=(-\infty)$

As we can see, $L_{left}$ and $L_{right}$ neither agree nor exist finitely and as such, the two sided limit $L$ doesn't exist.