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How do you compute the limit of #sin(7x)/sin(2x)# as #x->0#?

Answer 1

Charles Anctil

#lim_(x->0)frac sin(7x) sin(2x) =7/2#

Considering that:

#lim_(x->0) frac sin(alphax) (alphax) =1#

You can express:

#frac sin(7x) sin(2x) = 7x frac sin(7x) (7x) frac (2x) sin(2x) 1/(2x)#

#frac sin(7x) sin(2x) = 7/2 frac frac sin(7x) (7x) frac sin(2x) (2x) #

and then:

#lim_(x->0)frac sin(7x) sin(2x) = lim_(x->0)7/2 frac frac sin(7x) (7x) frac sin(2x) (2x) =7/2 *1/1 =7/2#

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Answer 2

Charles Arocha

To compute the limit of sin(7x)/sin(2x) as x approaches 0, we can use the concept of limits and trigonometric identities. By applying the limit properties and the fact that sin(x)/x approaches 1 as x approaches 0, we can simplify the expression.

Using the trigonometric identity sin(2x) = 2sin(x)cos(x), we can rewrite the expression as (sin(7x))/(2sin(x)cos(x)).

Next, we can cancel out the sin(x) term in the numerator and denominator, resulting in (sin(7x))/(2cos(x)).

Now, as x approaches 0, sin(7x) and cos(x) both approach 0. Therefore, the limit of (sin(7x))/(2cos(x)) as x approaches 0 is 0/2, which equals 0.

Hence, the limit of sin(7x)/sin(2x) 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 compute the limit of #sin(7x)/sin(2x)# as #x-&gt;0#?

How do you compute the limit of #sin(7x)/sin(2x)# as #x->0#?