Find display stylelim_{x to 0} dfrac{ tan(x)}{3x+ tan(x)} x→0 lim 3x+tan(x) tan(x) limit, start subscript, x, to, 0, end subscript, start fraction, tangent, (, x, ), divided by, 3, x, plus, tangent, (, x, ), end fraction

lim_(x→0)[tan(x)/(3x+ tan(x))]

= 1/4

Step-by-step explanation:

lim_(x→0)[tan(x)/(3x+ tan(x))]

Dividing both the numerator and denominator by tan(x), we have the limit to be

= lim_(x→0)[1/(3x/tan(x) + 1)]

= lim_(x→0)[1/(3xcos(x)/sin(x) + 1)]

= lim_(x→0)[sin(x)/(3xcos(x) + sin(x))]

Substituting x = 0, the limit gives an unwanted 0/0. So we apply L'Hopital's rule.

THE RULE STATES THAT

lim_(x→a)[f(x)/g(x)] = lim_(x→a)[f'(x)/g'(x)]

If lim_(x→a)[f(x)/g(x)] = 0/0,

it is reasonable to use

lim_(x→a)[f'(x)/g'(x)]

So

lim_(x→0)[sin(x)/(3xcos(x) + sin(x))]

= lim_(x→0)[cos(x)/(3cos(x) - 3xsin(x) + cos(x))]

= 1/(3 - 0 + 1)

= 1/4

differnt varibles efect the range

step-by-step explanation: