Find the limit as x approaches infinity of sin(2x)/x
The answer is 0. How can I show that this is an indeterminate form so I can use L'Hopital's? And even if I use L'Hopital's, it comes out to be 2cos(2x), which doesn't help if I replace x with infinity.
You can't use L'Hopital's rule on this. Use the fact that sin 2x is bounded between -1 and +1 and the x in the denominator goes to infinity. The absolute value of the limit is equal to orless than the limit of 1/x as x-> infinity, which is zero
To show that the given limit is an indeterminate form, you can rewrite it as:
lim(x→∞) sin(2x)/x = lim(x→∞) 2sin(x)cos(x)/x
Here, both sin(x) and cos(x) oscillate between -1 and 1 as x approaches infinity. However, the division by x in the denominator causes the overall value to diminish as x gets larger. As a result, we have a situation where the numerator oscillates between -2 and 2 while the denominator increases without bound.
To evaluate the limit, you can indeed use L'Hopital's rule by differentiating the numerator and denominator separately with respect to x. Applying it once, we get:
lim(x→∞) [d(sin(2x))/dx] / [d(x)/dx]
= lim(x→∞) 2cos(2x) / 1
= 2cos(2x)
At first glance, it may appear that replacing x with infinity does not yield a definitive value. However, it is important to remember that this is not the value of the limit. The limit is determined by the behavior of the function as x approaches infinity, rather than its value at infinity.
Since cos(2x) oscillates between -1 and 1 as x approaches infinity, the limit of 2cos(2x) as x approaches infinity does not exist. Therefore, the original limit, lim(x→∞) sin(2x)/x, is undefined.