find the limit as x approaches 0 of (tan7x)/(3x)?

Answer 1

=7/3

Answer 2

Use L'Hopital's rule:

d/dx (tan7x) = 7sec^2(7x) = 7/cos^2(7x)

d/dx (3x) = 3

Therefore, the limit as x approaches to 0 = (7/cos^2(0))/3 = (7/1)/3 = 7/3

Answer 3

Use l'hopitals rule. Since tan(7x)/3x is indeterminate (0/0) for x=0, you can differentiate both the denominator and the numerator to give the same result.

Lim tan(7x) / 3x = Lim 7 sec^2 (7x) / 3

Now simply plug zero into it:

Lim 7 sec^2 (7x) / 3 = 7/3 sec^2 (0) = 7/3

Answer 4

In general, for every real a <>0, lim x --> 0 tan(ax)/x = lim x --> 0 (sin(ax)/cos(ax))/x = lim x --> 0 a * sin(ax)/(ax) *1/cos(ax)

We know lim u --> 0 sin(u)/u =1. Hence, lim x -->0 sin(ax)/(ax) = 1. And since the cossine is a continuous function, it follows that lim x --> 0 cos(ax) = cos(0) =1. So, by the properties of limits,

lim x --> 0 tan(ax)/x = a * 1 * 1 = a

In your case, we have lim x --> 0 1(/3) tan(7x)/x = 1/3 * 7 = 7/3

Answer 5

lim x->0 tan7x/3x
= lim x->0 (sin7x/3x) (1/cos7x)
= lim x->0 (sin7x/3x) lim x->0 (1/cos7x) *both limit exists
= lim x->0 (sin7x/7x) 7/3
= 7/3

notes:
lim x->0 (1/cos7x) = 1
lim x->0 (sin7x/7x) = 1

Answer 6

Does not exist. I don't know what tan0 is because I have been out of high school for a while, but I do not that you cannot have a denominator = 0, therefore the fraction is undefined, and the limit does not exist.