How do you find the limit of # (tan2x)/(5x)# as x approaches 0?

Question

Emily Ammerman · Accepted Answer

#lim_{x->0}tan(2x)/(5x) = 2/5#

#tan(2x)/(5x)=sin(2x)/(5x cos(2x)) = ((sin(2x))/(2x))(2/(5 cos(2x)))#

so

#lim_{x->0}tan(2x)/(5x) = lim_{x->0} ((sin(2x))/(2x)) xx lim_{x->0}(2/(5 cos(2x)))#

Finally

#lim_{x->0}tan(2x)/(5x) = 1 xx 2/5 = 2/5#

Christopher Agresta · Answer

undefined To find the limit of (tan2x)/(5x) as x approaches 0, we can use the limit definition. By applying the limit definition, we can simplify the expression and evaluate the limit.

First, we can rewrite tan2x as sin2x/cos2x.

Next, we can rewrite the expression as (sin2x)/(cos2x * 5x).

Now, we can evaluate the limit by considering the limits of the numerator and denominator separately.

As x approaches 0, sin2x approaches 0, and cos2x approaches 1.

Therefore, the limit of the numerator is 0.

For the denominator, as x approaches 0, cos2x * 5x approaches 5 * 0 = 0.

Therefore, the limit of the denominator is 0.

Since the limit of the numerator and denominator are both 0, we can apply L'Hopital's rule.

By differentiating the numerator and denominator, we get (2cos2x)/(5).

Now, we can evaluate the limit of (2cos2x)/(5) as x approaches 0.

As x approaches 0, cos2x approaches 1.

Therefore, the limit of (2cos2x)/(5) is (2 * 1)/(5) = 2/5.

Hence, the limit of (tan2x)/(5x) as x approaches 0 is 2/5.