# Taylor's theorem and evaluating limits when x goes to infinity

In this problem I have to evaluate the following limit $$\lim_{x \to \infty} \left(\left(x^3-x^2+\frac{x}{2}\right)e^{1/x} - \sqrt{x^6+1}\right)$$ using the Taylor's formula with the Peano form of the remainder.
In class I have seen examples of how to handle limits using Taylor series expansion when $$x$$ goes to $$0$$, but I am not sure how to solve something like this. I tried substituting $$t=1/x$$, which led to $$\lim_{t \to 0^+}\frac{(2-2t+t^2)e^t-2\sqrt{1+t^6}}{2t^3}.$$ Now it looks a little bit better but I got stuck at this point. Is this approach correct? If so, how do I proceed?

• It is correct. Now you have to expand the numerator, at least up to order $3$. Commented Jan 23, 2021 at 21:28

$$e^t \left(t^2-2 t+2\right)=2+\frac{t^3}{3}+O\left(t^4\right)$$ $$2\sqrt{t^6+1}=2+O\left(t^4\right)$$ thus $$\frac{e^t \left(t^2-2 t+2\right)-2 \sqrt{t^6+1}}{2 t^3}\sim \frac{2+\frac{t^3}{3}-2}{2 t^3},\text{ as }x\to 0^+$$ Therefore $$\underset{t\to 0^+}{\text{lim}}\frac{e^t \left(t^2-2 t+2\right)-2 \sqrt{t^6+1}}{2 t^3}=\frac{1}{6}$$