Show $R(x)=o(x^3)$

Question

I got $$R(x)=4! \, x^4 \int _0^{\infty} \frac{1}{(1+xt)^{5}}e^{-t} \, \, dt$$ is this correct?

I have no idea what to do for the last part of ii

Answer 1

Need to show $$\lim_{x \rightarrow 0} \frac{R(x)}{x^3}=4! \lim_{x \rightarrow 0} x \int_0^{\infty} \frac{e^{-t}}{(1+xt)^5}\, dt=0$$ so we need to show $$ A= \lim_{x \rightarrow 0} x \int_0^{\infty} \frac{e^{-t}}{(1+xt)^5}\, dt=0 $$ Then $$0 \le A \le x \int_0^{\infty} {e^{-t}} \, dt = x $$ so $$0 \le A \le x$$ Then since $$\lim_{x \rightarrow 0^+}0 =\lim_{x \rightarrow 0^+} x =0$$ then by squeeze rule, $$\lim_{x \rightarrow 0^+} A=0$$