lim of ((-5+h)^2-25)/(h) as x approaches 0

Can someone explain to me step by step how this equals -10...??

Question

lim of ((-5+h)^2-25)/(h) as x approaches 0 


Can someone explain to me step by step how this equals -10...??

anonymous · Answer

$$\frac{ (-5+h)^{2}-25 }{ h }=\frac{ 25-10h+h^{2}-25 }{ h }=-10+h$$

campbell_st · Answer

well start with writing it correctly 

$$\lim_{h \rightarrow 0}\frac{( -5 + h)^2 - 25}{h}$$

simplify the numerator 

$$\lim_{h \rightarrow 0}\frac{25 - 10h + h^2 - 25}{h} = \lim_{h \rightarrow 0}\frac{-10h + h^2}{h}$$

remove to common factor 

$$\lim_{h \rightarrow 0} \frac{h(-10 + h)}{h} =\lim_{h \rightarrow 0} -10 + h$$

now substitute h = 0 into the limit

-10 + 0 = -10

anonymous · Answer

Thanks guys! that helped a lot... i think i forgot to simplify and instead went right ahead and plugged in the 0 giving me a DNE