Prove $\lim_{r\longrightarrow0^+} \left[\int_{-1}^{-r} \frac{f(x)}x \;\mathrm{d}x + \int_{r}^1 \frac{f(x)}x \;\mathrm{d}x\right]$ exists. 
Suppose $f$ is continuous over $[-1, 1]$ and differentiable at $0$. Prove the following limit exists,
  $$\lim_{r\longrightarrow0^+} \left[\int_{-1}^{-r} \frac{f(x)}x \;\mathrm{d}x + \int_{r}^1 \frac{f(x)}x \;\mathrm{d}x\right]$$

I encountered this problem in a test. I tried to change the limits by substitution, and then got a common expression for both integrals by change of variable. However, I have been confused over what the problem is asking.
 A: Because it equals 
$$\lim_{r\to 0+} \int_r^1 \frac{f(x)-f(-x)}{x} \mathrm{d}x=\lim_{r\to 0+} \int_r^1 \big(\frac{f(x)-f(0)}{x}+\frac{f(0)-f(-x)}{x}\big)\mathrm{d}x$$ and the function you integrate now is continuous and can be continuously extended to $r=0$ (because $f$ is differentiable in $0$).
A: Here is another proof of the property of the original poster.
Theorem :
If $\;f:[-1,1]\to\mathbb{R}\;$ is a continuous function over $\;[-1, 1]\;$ and is differentiable at $\;x=0\;,\;$ then the following limit exists and is finite ,
$\displaystyle\lim_{r\to0^+} \left[\int_{-1}^{-r} \frac{f(x)}x \;\mathrm{d}x + \int_{r}^1 \frac{f(x)}x \;\mathrm{d}x\right]\;.$

Proof :
Let $\;\varphi(x)\;$ be the function $\;\varphi:[-1,1]\to\mathbb{R}\;$ defined by
$\varphi(x)=\begin{cases}\dfrac{f(x)-f(0)}x\quad \text{for all }\;x\in [-1,1]\setminus\{0\}\\f’(0)\qquad\quad\;\;\text{for }\;x=0\qquad\qquad\qquad\;\;.\end{cases}$
$\varphi(x)\;$ is a continuous function over $\;[-1,1]\;,\;$ hence it has an antiderivative over $\;[-1,1]\;$ that is there exists a function
$\Phi(x):[-1,1]\to\mathbb{R}\;$ differentiable over $\;[-1,1]\;$ such that
$\Phi’(x)=\varphi(x)\quad\forall x\in [-1,1]\;.$
Moreover, it results that
$f(x)=f(0)+x\varphi(x)\quad\forall x\in [-1,1]\;\;,$
$\displaystyle\int\frac{f(x)}x\mathrm{d}x=\int\left[\frac{f(0)}x+\varphi(x)\right]\!\mathrm{d}x=f(0)\ln|x|+\Phi(x)+\text{c}\;,$
$\begin{align}
&\!\displaystyle\int_{-1}^{-r}\frac{f(x)}x\mathrm{d}x+\int_r^1\frac{f(x)}x\mathrm{d}x=\\
&\;\;=\bigg[f(0)\ln|x|+\Phi(x)\bigg]_{-1}^{-r}+\bigg[f(0)\ln|x|+\Phi(x)\bigg]_r^1=\\
&\;\;=f(0)\ln r+\Phi(-r)-\Phi(-1)+\Phi(1)-f(0)\ln r-\Phi(r)=\\
&\;\;=\Phi(-r)-\Phi(-1)+\Phi(1)-\Phi(r)\;\;,
\end{align}$
for all $\;r\in\,]0,1]\;.$
Furthermore,
$\begin{align}
\displaystyle\lim_{r\to0^+}&\left[\int_{-1}^{-r}\frac{f(x)}x\mathrm{d}x+\int_r^1\frac{f(x)}x\mathrm{d}x\right]=\\
&=\lim_{r\to0^+}\bigg[\Phi(-r)-\Phi(-1)+\Phi(1)-\Phi(r)\bigg]=\\
&=\Phi(1)-\Phi(-1)\;\;,
\end{align}$
because $\;\Phi(x)\;$ is continuous over $\;[-1,1]\;.$
Hence there exists and is finite the limit
$\displaystyle\lim_{r\to0^+}\left[\int_{-1}^{-r}\frac{f(x)}x\mathrm{d}x+\int_r^1\frac{f(x)}x\mathrm{d}x\right]=\Phi(1)-\Phi(-1)\;.$
