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How do you find the limit of #sin^2x/x# as #x->0#?

Answer 1

The limit is #0#.

Consider the well-known limit #lim_(x->0) sinx/x = 1#. We will need to rewrite this expression in order to use this property.

#=lim_(x-> 0)( sinx/x xx sinx)#

#=lim_(x->0) sinx/x xx lim_(x->0)sinx#

#=1 xx 0#

#=0#

Hopefully this helps!

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Answer 2

Christopher Allers

To find the limit of sin^2x/x as x approaches 0, we can use the concept of limits and trigonometric identities. By applying the limit definition, we can rewrite the expression as (sinx/x) * sinx. Since the limit of sinx/x as x approaches 0 is 1, the limit of sin^2x/x as x approaches 0 is equal to 1 * sin(0), which simplifies to 0. Therefore, the limit of sin^2x/x as x approaches 0 is 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.

How do you find the limit of #sin^2x/x# as #x-&gt;0#?

How do you find the limit of #sin^2x/x# as #x->0#?