Find the limit
lim tan^2(7x)/9x
x-> 0
recall that tanx/x = sinx/x = 1
as x -> 0
so, this limit is
7/9 * tan(7x)/(7x) * tan(7x)
-> 7/9 * 1 * 0 = 0
To find the limit of the given expression as x approaches 0, we can use basic trigonometric properties and algebraic manipulation.
Let's simplify the expression first.
lim (x->0) tan^2(7x)/9x
Since tan(0) = 0, we can substitute 0 in place of x in the expression:
lim (x->0) tan^2(7x)/9x = tan^2(0)/9(0)
Using the property that tan(0) = 0:
0^2/9(0) = 0/0
Now, we have an indeterminate form 0/0. To further evaluate this limit, we can apply L'Hôpital's rule, which states that if we have an indeterminate form of the type 0/0 or ∞/∞, we can take the derivative of the numerator and denominator and then calculate the limit again.
Differentiating the numerator and denominator:
lim (x->0) [d/dx(tan^2(7x))] / [d/dx(9x)]
= lim (x->0) [2tan(7x)*sec^2(7x)*7] / [9]
Now, substitute x = 0 into the expression:
= 2tan(0)*sec^2(0)*7 / 9
Since tan(0) = 0 and sec(0) = 1:
= 2 * 1 * 7 / 9
Simplifying:
= 14/9
Therefore, the limit of the given expression as x approaches 0 is 14/9.