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How do you find the limit of #(sin3x)/x# as #x->0#?

Answer 1

Cameron Alexander

The limit is #3#.

#lim_(xrarr0)(sin3x)/x = lim_(xrarr0)(3sin(3x))/((3x)) #

# = 3 lim_(xrarr0)(sin(3x))/((3x)) #

Now use #lim_(theta rarr0) sintheta / theta = 1# with #theta = 3x# to get

# = 3(1)=3#

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Answer 2

Ezekiel Alexander

To find the limit of (sin3x)/x 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 with respect to x.

Differentiating sin3x with respect to x gives us 3cos3x, and differentiating x with respect to x gives us 1.

Now, we can evaluate the limit of (3cos3x)/1 as x approaches 0. Plugging in x=0 into the expression gives us 3cos(0)/1, which simplifies to 3(1)/1 = 3.

Therefore, the limit of (sin3x)/x as x approaches 0 is equal to 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.

How do you find the limit of #(sin3x)/x# as #x-&gt;0#?

How do you find the limit of #(sin3x)/x# as #x->0#?