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@@ -67,19 +67,27 @@ $$ \frac{1}{x} - \frac{1}{x+1} = \frac{x + 1 - x}{x^2 + x} = \frac{1}{x^2 + x}$$
 
 \section*{Question 3}
 \subsection*{Point a)}
-First we point out that:
-$$ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \stackrel{k := -h}{=} \lim_{k \to 0} \frac{f(x - (-k)) - f(x)}{-k} = $$
-$$ =  \lim_{k \to 0} \frac{f(x) - f(x + k)}{k} = - \lim_{h \to 0} \frac{f(x) - f(x - h)}{h} $$ 
+
+Consider the Taylor expansion with $a=x$ of $f(x+h)$ and $f(x-h)$:
+
+$$f(x+h) \geq f(x) + \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 + 
+\frac{f'''(x)}{6}\cdot h^3$$
+
+  $$f(x-h) \geq f(x) - \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 -
+\frac{f'''(x)}{6}\cdot h^3 $$ 
  
-Then, we find an equivalent way to represent $f'(x)$:
-$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 2 \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} =$$
-$$ = \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} + \left(- \lim_{h \to 0} \frac{f(x) - f(x - h)}{2h}\right) =  \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} $$
+Then, we can derive that:
 
-Then we consider the \textit{epsilon-delta} definition of limits for the last limit:
-$$ \forall \epsilon > 0 \exists \delta > 0 | \forall h > 0, \text{if } 0 < h < \delta \Rightarrow$$
-$$\left| \frac{f(x + h) - f(x - h)}{2h} - f'(x)\right| = \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| < \epsilon$$
+$$\frac{f(x + h) - f(x - h)}{2h} \geq 
+\frac{1}{2h} \cdot 
+\left(2hf'(x) + \frac{2h^3f'''(x)}{6} \right) =
+f'(x) + \frac{h^2f'''(x)}{6}
+$$
 
-I GIVE UP :(
+So:
+
+$$ \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| \leq \left|f'(x) - \left(f'(x) + \frac{h^2f'''(x)}{6}\right)
+\right| = \frac{h^2|f'''(x)|}{6}$$
 
 \section*{Question 4}
 \subsection*{Point a)}