Question

The value of ${lim}_{n\to \infty }\left(\frac{x^4\sin \frac{1}{x}+x^3}{1+\left|x^3\right|}\right)$

Answer 1

Let

$x=\frac{1}{t}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\left(\frac{1}{t^4}\sin t+\frac{1}{t^3}\right)}{1+\frac{1}{t^3}}$

Now,

$L.H.S=\underset{h\to 0^{+}}{lim}\frac{\left(\frac{1}{h^4}\sinh +\frac{1}{h^3}\right)}{\left(1+\frac{1}{h^3}\right)}$

$=\underset{h\to 0^{+}}{lim}\frac{\sinh +h}{h^4+h}=\underset{h\to 0^{+}}{lim}\frac{\cosh +1}{4h^3+1}\left[byL-hospitalrule\right]$

$=\frac{1+0}{0+1}=2$

Now,

$R.H.S=\underset{h\to 0^{-}}{lim}\frac{\left(\frac{1}{h^4}\left(\sin \left(-h\right)\right)+\frac{1}{{\left(-h\right)}^3}\right)}{\left(1-\frac{1}{\left(-h^3\right)}\right)}$

$=\underset{h\to 0^{-}}{lim}\frac{\frac{-1}{h^4}\sinh -\frac{1}{h^3}}{1+\frac{1}{h^3}}=\underset{h\to 0^{-}}{lim}\frac{-\sinh -h}{h^4+h}$

$=\underset{h\to 0^{-}}{lim}\frac{-\cosh -1}{4h^3+1}\left(byL-Hospitalrule\right)$

$=\frac{-1-1}{0+1}=-2$

$So,L.H.S\ne R.H.S$

So, limit does not exist.