Question
Is the integral $$\int^1_{-1} \left(\frac 1 {x^5} + x^2\right)\ dx$$ convergent or divergent?
My working
Clearly, $$\int^1_{-1} x^2\ dx$$ is always convergent, so we check $$\int^1_{-1} \frac 1 {x^5}\ dx,$$ where $$\begin{aligned} \int^1_{-1} \frac 1 {x^5}\ dx & = \int^0_{-1} \frac 1 {x^5}\ dx + \int^1_0 \frac 1 {x^5}\ dx\\[2 mm] & = \lim\limits_{c \to 0^-} \left[-\frac 1 {4x^4}\right]^c_{-1} + \lim\limits_{c \to 0^+} \left[-\frac 1 {4x^4}\right]^1_c\\[2 mm] & = \lim\limits_{c \to 0^-} \left(-\frac 1 {4c^4}\right) + \lim\limits_{c \to 0^+} \left(\frac 1 {4c^4}\right)\\[2 mm] & = -\infty + \infty, \end{aligned}$$ which is indeterminate, so I concluded that the integral is divergent. However, the solution goes from $$\lim\limits_{c \to 0^-} \left(-\frac 1 {4c^4}\right) + \lim\limits_{c \to 0^+} \left(\frac 1 {4c^4}\right)$$ to $0$ and concludes that the integral is convergent.
I am not sure where I have gone wrong, so any intuitive explanations will be greatly appreciated!