What is the limit as x approaches infinity of 4xsin(1/x)?
Can you also show me how you get the answer?
Thank you so much!!
lim (x --> inf) 4x * sin(1/x)
Let t = 1/x as x --> inf, t --> 0
lim (t --> 0) 4 (1/t) * sin(t)
lim (t --> 0) 4 * sin(t)/t
lim (t--> 0) 4 * lim (t --> 0) sin(t) / t
= 4 * 1
= 4
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To find the limit as x approaches infinity of 4xsin(1/x), we can use the concept of L'Hôpital's Rule and the properties of limits.
First, let's write the expression as a fraction:
4xsin(1/x) = 4sin(1/x) / (1/x)
Now, let's find the limit of the numerator and denominator separately as x approaches infinity.
For the numerator, the sine function is bounded between -1 and 1. So, as x approaches infinity, sin(1/x) will be squeezed between -1 and 1. Therefore, the limit of sin(1/x) as x approaches infinity is 0.
For the denominator, 1/x approaches 0 as x approaches infinity.
Now, we have the form 0/0, a limit that is indeterminate. To resolve this, we can apply L'Hôpital's Rule. This rule states that if we have a limit of the form 0/0 or ∞/∞, we can take the derivative of the numerator and denominator individually and evaluate the limit again.
Taking the derivative of the numerator:
d/dx (4sin(1/x)) = 4cos(1/x) * (-1/x^2)
Taking the derivative of the denominator:
d/dx (1/x) = -1/x^2
Now, we can evaluate the limit again. As x approaches infinity:
lim(x->∞) [4cos(1/x) * (-1/x^2)] / (-1/x^2)
Simplifying the expression:
lim(x->∞) (-4cos(1/x))
Now, as x approaches infinity, cos(1/x) oscillates between -1 and 1 but never settles on a single value. So, lim(x->∞) cos(1/x) does not exist.
Therefore, the overall limit as x approaches infinity of 4xsin(1/x) is also undefined.
In summary, the limit as x approaches infinity of 4xsin(1/x) does not exist.