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Given #sin x/ x^2# how do you find the limit as x approaches 0?

Answer 1

Luke Alvarez

You should use the fact that #lim sinx/x = 1# as #x# approaches 0

You can modify the expression as follows:

#sinx/x^2 = sinx/x * 1/x #, and then taking the limits:

#lim sinx/x = 1#, and

# lim 1/x =oo #, so the end result is:

#lim sinx/x^2 = lim sinx/x * lim 1/x = 1 * oo= oo#

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Answer 2

Isabella Ackerman

To find the limit as x approaches 0 for sin(x)/x^2, we can use L'Hôpital's Rule. By applying this rule, we differentiate both the numerator and denominator separately with respect to x.

The derivative of sin(x) is cos(x), and the derivative of x^2 is 2x.

Taking the limit as x approaches 0, we have:

lim(x→0) sin(x)/x^2 = lim(x→0) cos(x)/2x

Now, we can evaluate this new limit. By substituting x = 0 into the expression, we get an indeterminate form of 0/0.

To resolve this, we can again apply L'Hôpital's Rule. Differentiating the numerator and denominator once more, we have:

lim(x→0) cos(x)/2x = lim(x→0) -sin(x)/2

Now, substituting x = 0 into the expression, we get -sin(0)/2 = 0/2 = 0.

Therefore, the limit as x approaches 0 for sin(x)/x^2 is equal to 0.

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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.