Limit x to o: $\lim_{h\to 0}=-\frac{1}{5}$

[karush]

$#46p60$
$$\lim_{{h}\to{0}}\frac{\frac{1}{5+h}-\frac{1}{5}}{h}=-\frac{1}{5}$$

Multiply numerator and denomator by $\frac{1}{h}$
$$\frac{-1}{5h+25}$$
$$h\to 0$$ thus $\frac{-1}{25}$

I hope anyway

 

[Guest2]

Correct! Alternatively, you might recognize this as the derivative of 1/x evaluated at x = 5, i.e.

$\displaystyle \lim_{{h}\to{0}}\frac{\frac{1}{5+h}-\frac{1}{5}}{h} = \frac{d}{dx} \left(\frac{1}{x}\right)\bigg|_{x=5} = -\frac{1}{x^2}\bigg|_{x=5} = -\frac{1}{25}.$

 

[karush]

https://dl.orangedox.com/DllM3xBiyI8YRHbmhY