How do you solve the limit as h approaches 0 of #[34/(35+h) - (34/35)]/h#?

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Answer 1 · 2015-06-26T14:05:46

#lim_(x->0)[34/(35+x) - 34/35]/x =-34/1225 ~~ -0.028#.

#lim_(x->0)f(x)/g(x)= lim_(x->0)[34/(35+x) - 34/35]/x = ''(0/0)''#, which is undefined.

We have #f(x) = 34/(35+x) - 34/35# and #g(x) = x#.

Since #lim_(x->0)f(x)=lim_(x->0)g(x)=0#, we can use L'Hospital's Rule :

#lim_(x->0)f(x)/g(x)=lim_(x->0)(f'(x))/(g'(x)) =lim_(x->0)([34/(35+x) - 34/35]')/((x)')#

#=lim_(x->0)((34/(35+x))')/((x)')#

#=lim_(x->0)((-34)/(35+x)^2)/1 = lim_(x->0)(-34)/(35+x)^2 = -34/35^2 = -34/1225 ~~ -0.028#.

Answer 2 · 2015-06-26T19:14:08

Rewrite (simplify) the expression.

Start by writing the numerator as a single ratio:

#34/(35+h) - 34/35 = ((34)(35)-(34)(35+h))/((35+h)(35))#

# =(-34h)/(35(35+h))#

So the big ratio becomes:

#(34/(35+h) - 34/35)/h = (((34)(35)-(34)(35+h))/((35+h)(35)))/(h/1)#

# =(-34h)/(35(35+h))*1/h#

# = (-34)/(35(35+h))#

For the limit, we get:

#lim_(hrarr0)(34/(35+h) - 34/35)/h = lim_(hrarr0)(-34)/(35(35+h)) = (-34)/35^2#