Solve lim sin2x using any method.
×>0 -----
5x
I am unaware of what my first step is supposed to be, I know that if I plug in 0 to x, I will receive an answer of 0/0, and that makes my answer indeterminate. So what happens after that?
You know that
lim sinu/u = 1, so
sin2x/5x = sin2x/2x * 2/5
To solve the limit lim sin(2x)/(5x) as x approaches 0, you can use L'Hôpital's rule, which allows you to differentiate the numerator and denominator separately.
Here are the steps to apply L'Hôpital's rule in this case:
Step 1: Plug in x = 0 into the expression to see if it results in an indeterminate form. In this case, you correctly identified that you get 0/0, which is an indeterminate form.
Step 2: Differentiate the numerator and denominator separately. The derivative of sin(2x) is 2*cos(2x), and the derivative of 5x is simply 5.
Step 3: Now, calculate the limit of the derivatives of the numerator and denominator separately. As x approaches 0, the derivative 2*cos(2x) approaches 2*cos(0) = 2, and the derivative 5 approaches 5. Hence, the limit of the derivatives is 2/5.
Step 4: Since the limit of the derivatives is not indeterminate anymore, you can conclude that the limit of the original function is the same as the limit of the derivatives, which is 2/5.
Therefore, lim sin(2x)/(5x) as x approaches 0 is equal to 2/5.