http://people.hofstra.edu/Stefan_Waner/proofs/quotientruleproof.html

this is the website show the proves of quotient rule. I = don't understand the denominator at last step, does g(x+h)g(x) = [g(x)]² ?

Question

http://people.hofstra.edu/Stefan_Waner/proofs/quotientruleproof.html

this is the website show the proves of quotient rule. I = don't understand the denominator at last step, does g(x+h)g(x) = [g(x)]²  ?

anonymous · Answer

is it same as g'(x) x g(x) ?

anonymous · Answer

i mean is it same as [g'(x)][ g(x)]   ?

anonymous · Answer

as h ---> 0  it is valid to say g(x+h)g(x) = [g(x)]²   . h becomes so small that it can be neglected

anonymous · Answer

so the result using this formula is approximate form?

anonymous · Answer

Nice jimmy!

anonymous · Answer

not really approximate the value of becomes infinitessimally small

anonymous · Answer

ok, thanks u, i get it ^^

anonymous · Answer

you can assert that 
$$\lim_{h\rightarrow 0}g(x+h)=g(x+0)=g(x)$$ precisely because you are assuming that g is differentiable, and therefore g must be continuous. it is the continuity of g that allows you to take the limit inside the function, otherwise this is not true

anonymous · Answer

just thought i would mention it. in general you cannot say that 
$$\lim_{h\rightarrow 0}g(x+h)=g(x)$$