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

Question

2585 views around the world

You can reuse this answer
Creative Commons License

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:

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

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