Given f(x) = cases arctan (1+ 1/x), x ≠ 0,, [6pt] 0, x =

Given $f(x) = \begin{cases} \arctan\!\left(1+\dfrac{1}{x}\right), & x \neq 0, \\[6pt] 0, & x = 0, \end{cases}$ find (a) $f'_{-}(0)$ and $f'_{+}(0)$; (b) $\lim_{x\to 0^{+}}f'(x)$ and $\lim_{x\to 0^{-}}f'(x)$. Does $f'(0)$ exist?

Step 1: Part (a): the right-hand derivative $f'_{+}(0)$

For $\Delta x \gt 0$, using $f(0) = 0$: \[\frac{f(\Delta x)-f(0)}{\Delta x} = \frac{\arctan\!\left(1+\dfrac{1}{\Delta x}\right)}{\Delta x}.\]

As $\Delta x \to 0^{+}$ the argument $1+\dfrac{1}{\Delta x} \to +\infty$, so the numerator tends to the positive constant $\dfrac{\pi}{2}$ while the denominator tends to $0$ through positive values:

\[ f'_{+}(0) = \lim_{\Delta x \to 0^{+}} \frac{\arctan\!\left(1+\frac{1}{\Delta x}\right)}{\Delta x} = \frac{\pi/2}{0^{+}} = +\infty. \]

Step 2: Part (a): the left-hand derivative $f'_{-}(0)$

For $\Delta x \lt 0$ the argument $1+\dfrac{1}{\Delta x} \to -\infty$, so the numerator tends to $-\dfrac{\pi}{2}$ while the denominator tends to $0$ through negative values. A negative number divided by a negative number is positive:

\[ f'_{-}(0) = \lim_{\Delta x \to 0^{-}} \frac{\arctan\!\left(1+\frac{1}{\Delta x}\right)}{\Delta x} = \frac{-\pi/2}{0^{-}} = +\infty. \]

Step 3: Part (b): compute $f'(x)$ for $x \neq 0$

Apply the chain rule with $u = 1+\dfrac{1}{x}$, so that $u' = -\dfrac{1}{x^{2}}$ and $\bigl(\arctan u\bigr)' = \dfrac{1}{1+u^{2}}$: \[f'(x) = \frac{1}{1+\left(1+\dfrac{1}{x}\right)^{2}}\cdot\left(-\frac{1}{x^{2}}\right) = -\frac{1}{x^{2}\left[1+\left(1+\dfrac{1}{x}\right)^{2}\right]}.\]

Multiplying the bracket through by $x^{2}$ simplifies the expression considerably: $f'(x) = -\frac{1}{x^{2} + (x+1)^{2}}$.

Step 4: Part (b): the one-sided limits of $f'$

The simplified formula is a continuous function of $x$ near the origin, with denominator $x^{2}+(x+1)^{2} \to 1$: \[\lim_{x\to 0^{+}} f'(x) = -\frac{1}{0+1} = -1, \qquad \lim_{x\to 0^{-}} f'(x) = -\frac{1}{0+1} = -1.\]

Step 5: Conclusion on $f'(0)$

The one-sided derivatives at the origin are infinite, so $f'(0)$ does not exist as a finite number. Note the apparent paradox: $f'(x) \to -1$ from both sides, yet the difference quotient at the origin diverges to $+\infty$.

Step 6: Explanation of the apparent paradox

There is no contradiction, because $f$ is not continuous at the origin. Indeed

\[\lim_{x\to 0^{+}} f(x) = \arctan(+\infty) = \frac{\pi}{2}, \qquad \lim_{x\to 0^{-}} f(x) = \arctan(-\infty) = -\frac{\pi}{2}\]

\[f(0) = 0 \neq \pm\frac{\pi}{2} \;\Longrightarrow\; f \text{ has a jump discontinuity at } x = 0\]

The value $f(0) = 0$ sits strictly between the two one-sided limits, so approaching the origin from either side the numerator of the difference quotient tends to a non-zero number of the same sign as $\Delta x$. That is exactly why both one-sided derivatives blow up to $+\infty$. The theorem that $f'(x) \to L$ forces $f'(0) = L$ requires continuity at the point, which fails here.