Try to rationalize this!

Substituting $x = 1/h$ . we have $h \to 0+$ as $x \to \infty$ .

So rewriting the expression we have

$(\frac{(1+h)^{2015}}{h^{2015}} + \frac{2013}{h^{2000}})^{\frac{1}{2015}} - \frac{1}{h}$

so we have:-

$lim_{h\to0^{+}}$ $\frac{((1+h)^{2015} + 2013h^{5})^{\frac{1}{2015}} - 1}{h}$ which is $(\frac{0}{0})$ form

So By L'Hospital Rule:-

P.S:- We can also make an approximation at this step as $((1+h)^{2015} + 2013h^{5})^{\frac{1}{2015}}$ tends to $1+h$ as $h \to 0^{+}$

we have our answer as $1$ .