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How do you evaluate the limit #sin(5x)/x# as x approaches #0#?

Answer 1

Elijah Abram

Use #lim_(thetararr0)sintheta/theta = 1# and some other tools.

#lim_(xrarr0)sin(5x)/x#

We'd like to use the limit mentioned above, so we need to have #theta = 5x#

To get #5x# in the denominator, we'll multiply by #5/5#

#lim_(xrarr0)sin(5x)/x = lim_(xrarr0)(5sin(5x))/(5x) #

Now factor the #5# in the numerator outside the limit.

# = 5lim_(xrarr0)sin(5x)/(5x) #

As #xrarr0#, #5xrarr0# so we have:

# = 5(1) = 5#

(The limit is #5#.)

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Answer 2

Cameron Alexander

To evaluate the limit of sin(5x)/x as x approaches 0, 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(5x) is 5cos(5x), and the derivative of x is 1.

Taking the limit of the differentiated function, we have 5cos(5x)/1, which simplifies to 5cos(5x).

Now, we can substitute x = 0 into the simplified expression, giving us 5cos(0). The cosine of 0 is equal to 1, so the final result is 5.

Therefore, the limit of sin(5x)/x as x approaches 0 is equal to 5.

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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.