# If function f is differentiable at c, simplify lim_(h->0)((f(c+h^2)-f(c))/h)?

Jun 25, 2018

${\lim}_{h \to 0} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = 0$

#### Explanation:

Multiply and divide by $h$:

${\lim}_{h \to 0} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = {\lim}_{h \to 0} h \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h} ^ 2\right)$

Substitute $\eta = {h}^{2}$, clearly when $h \to 0$ also $\eta \to 0$ and:

$h = \sqrt{\eta}$ for $h > 0$
$h = - \sqrt{\eta}$ for $h < 0$

Then:

${\lim}_{h \to {0}^{+}} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = {\lim}_{\eta \to 0} \sqrt{\eta} \left(\frac{f \left(c + \eta\right) - f \left(c\right)}{\eta}\right)$

As $f$ is differentiable in $c$:

${\lim}_{\eta \to 0} \left(\frac{f \left(c + \eta\right) - f \left(c\right)}{\eta}\right) = f ' \left(c\right)$

so:

${\lim}_{h \to {0}^{+}} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = {\lim}_{\eta \to 0} \sqrt{\eta} {\lim}_{\eta \to 0} \left(\frac{f \left(c + \eta\right) - f \left(c\right)}{\eta}\right) = f ' \left(c\right) {\lim}_{\eta \to 0} \sqrt{\eta} = 0$

and similarly:

${\lim}_{h \to {0}^{-}} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = - {\lim}_{\eta \to 0} \sqrt{\eta} {\lim}_{\eta \to 0} \left(\frac{f \left(c + \eta\right) - f \left(c\right)}{\eta}\right) = - f ' \left(c\right) {\lim}_{\eta \to 0} \sqrt{\eta} = 0$

Then:

${\lim}_{h \to 0} \left(\frac{f \left(c + {h}^{2}\right) - f \left(c\right)}{h}\right) = 0$