Find the limit as x->0 of (2-2cos(x))/(sin(5x))
Mathematically I got 2/5, but on the graph it appears to be 0.
Not sure how you got 2/5, but I got 0 as well.
Here's what I did:
Lim x->0 (2-2cos(x))/sin(5x)
both numerator and denominator evaluate to zero, so we can use d'Hôpital's rule:
=Lim x->0 (0+2sin(x))/5cos(5x)
This new expression evaluates to 0/5=0, which is therefore the answer.
Thanks. I made a really stupid error. You've been very helpful in answering my calc questions. I really appreciate it.
You're very welcome.
Keep up the good work. Calculus is best learned with lots of practice.
To find the limit as x approaches 0 of the given expression (2 - 2cos(x)) / sin(5x), we can use L'Hôpital's rule, which states that if we have an indeterminate form (such as 0/0 or ∞/∞), the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives under certain conditions.
Let's calculate the derivatives of the numerator and the denominator:
Taking the derivative of 2 - 2cos(x) with respect to x gives us 2sin(x).
Taking the derivative of sin(5x) with respect to x gives us 5cos(5x).
Now, let's rewrite the limit using these derivatives:
lim(x->0) (2 - 2cos(x)) / sin(5x)
= lim(x->0) (2sin(x)) / (5cos(5x))
Now evaluate this limit:
As x approaches 0, sin(x) approaches 0, and cos(5x) approaches 1.
Thus, the limit becomes:
= (2(0)) / (5(1))
= 0 / 5
= 0
Therefore, the limit as x approaches 0 of (2 - 2cos(x)) / sin(5x) is indeed 0, which matches what you observed on the graph.