How do you find the limit of #sinx / x# as x approaches infinity?

Answer 1

0

#lim_{x \to oo} sinx / x#
we know that #sin x \in [-1,1], qquad forall x#, ie it is a bounded function

so we can re-write the limit as

#sinx * \ lim_{x \to oo} 1 / x = sin x * 0 = 0#
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Answer 2

To find the limit of sin(x) / x as x approaches infinity, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate both the numerator and denominator with respect to x. The derivative of sin(x) is cos(x), and the derivative of x is 1.

Taking the limit of the derivative of sin(x) (which is cos(x)) divided by the derivative of x (which is 1) as x approaches infinity, we get:

lim(x→∞) cos(x) / 1

Since the cosine function oscillates between -1 and 1, it does not approach a specific value as x approaches infinity. Therefore, the limit of sin(x) / x as x approaches infinity is undefined.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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