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How do you find the limit of #(sin 2x)/(sin 3x)# as x approaches 0?

Answer 1

Christopher Allers

#L_(x->0)(sin2x)/(sin3x)=2/3#

We know that #L_(theta->0)(sintheta)/theta=1#

Hence #L_(x->0)(sin2x)/(sin3x)#

= #L_(x->0)[(sin2x)/(2x)((3x)/sin(3x))xx2/3]#

= #(L_(x->0)(sin2x)/(2x))/(L_(x->0)(sin3x)/(3x))xx2/3#

= #1/1xx2/3=2/3#

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Answer 2

Christopher Agresta

To find the limit of (sin 2x)/(sin 3x) as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately and then take the limit as x approaches 0 again.

Differentiating the numerator, we get 2cos 2x. Differentiating the denominator, we get 3cos 3x.

Taking the limit as x approaches 0 of the differentiated numerator and denominator, we have: lim(x→0) (2cos 2x)/(3cos 3x)

Now, we can substitute x = 0 into the expression: (2cos 2(0))/(3cos 3(0)) = (2cos 0)/(3cos 0) = 2/3

Therefore, the limit of (sin 2x)/(sin 3x) as x approaches 0 is 2/3.

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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.