Using the mean value theorem calculate this limit : $\lim_{x \rightarrow 0} \frac{arctan(x^2+x-1)+\frac{\pi}{4}}{x^2+3x}$

Question

Using the mean value theorem calculate this limit : $$\lim_{x \rightarrow 0} \frac{\arctan(x^2+x-1)+\frac{\pi}{4}}{x^2+3x}$$

I tried to apply Mean value theorem on $\arctan(x^2+x-1)$ but i didn't find the good result

Answer 1

Write $$ \arctan (x^2+x-1) + \frac{\pi}{4} = \arctan (-1+x+x^2)-\arctan(-1). $$ Therefore $$ \begin{align} \frac{\arctan (x^2+x-1) + \frac{\pi}{4}}{x^2+3x} &= \frac{\arctan (-1+x+x^2)-\arctan(-1)}{x^2+3x} \\ &= \frac{\arctan (-1+x+x^2)-\arctan(-1)}{x+x^2} \frac{x+x^2}{x^2+3x} \\ &\to \left( D \arctan (-1) \right) \cdot \frac{1}{3} = \frac{1}{6} \end{align} $$ as $x \to 0$, due to the definition of the derivative. Here $D$ denotes differentiation.

Answer 2

With $f(x)=\arctan(x^2+x-1)$, you want to find: $$\lim_{x \to 0} \frac{f(x)-f(0)}{x^2+3x} = \lim_{x \to 0} \frac{\frac{f(x)-f(0)}{x-0}}{x+3}$$ The Mean Value Theorem gives you a $c \in (0,x)$ such that: $$\frac{f(x)-f(0)}{x-0} = f'(c)$$ which means you also have: $$f_m'(x) \le \frac{f(x)-f(0)}{x-0} \le f_M'(x)$$ with $f_m'(x)$ the minimal value of $f'$ on $[0,x]$ and $f_M'(b)$ the maximal value.

The limit then follows by squeezing since: $$\lim_{x \to 0} \frac{f_m'(x)}{x+3} = \lim_{x \to 0} \frac{f_M'(x)}{x+3} = \frac{1}{6}$$