LIMITS: how can u find the limit for (sin2x)/x as x approaches 0?

Question

the answer is -1 but i don't know how, i asked this question before but i guess i didn't explain it right...you can't find this answer with a calculator, u can only find it algebraically, so to do that, u will have to simplify (sin2x)/x ...so this is where im stuck, i can't simplify it =[

thnx for ur help =]

Answer 1

There are a lot of ways to evaluate this limit. If you are in your first calculus course, you've just learned that \lim_{t \to 0} \sin(t) /t=1.

Using that, in this case we have that
\lim_{x \to 0} \sin(2x)/x=
2 \lim_{x \to 0} \sin(2x) / (2x) =
2 * 1=2.

Next semester, you'll learn L'Hospital's Rule. Basically, L'Hospital's rule says that if f(x)/g(x) is an indeterminate form of type 0/0 or infinity/infinity, then \lim_{x \to a} f(x)/g(x)=\lim_{x \to a} f'(x)/g'(x).

Applying L'Hospital's rule to this limit gives us

\lim_{x \to 0} \sin(2x)/x =
\lim_{x \to 0} 2 \cos(2x)/1=
2*1/1=2.

Another approach would be to divide the Maclaurin series for sin(2x) by x and then evaluate the limit of the sequence as x approaches 0. (The first term of the series Maclaurin series for sin(2x) /x is 2 which shows that the limit is 2.)

So, there are lots of ways to compute the limit. If you are in Calculus I, your teacher is expecting to see the first approach.

Good luck with your calculus!

Answer 2

No answer is not -1. This is a 0/0 indeterminate form and so you can use L'Hospital's rule. Take derivatives of both the numerator and the denominator, you get
lim 2cos2x/1 as x-->0
now if you plug in 0 you get a numerator of 2 and so the answer is 2.

Answer 3

I am using --> for approaches. I am using * for the multiplication symbol. Using l'Hopital's rule the limit is equal to the derivative of the numerator over the derivative of the denominator as x --> 0. The derivative of sin(2x) = 2cos(2x) As x--> 0 cos(2x) --> 1, because cos(0) = 1 2cos2x / 1 = (2 * 1)/1 = 2/1 = 2 <<<<<

Answer 4

Sin2x X

Answer 5

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u might know that lim sin(x)/x = 1 x-->0 so sin2x/x = 2sin(2x)/2x since sin(2x)/2x = 1 2sin(2x)/2x = 2*1 = 2 sin2x/x = 2 Im not very good in math.but i like to help u. contact me if u have any prob

Answer 6

Write sin 2x = 2 sin x cos x.
The limit of cos x as x goes to 0 is 1.
So we have the limit of 2 sin x/x as x goes to 0 = 2.
This last limit can be found in any calculus book,
though I don't like the usual proof by picture.

Answer 7

Use L'Hospital's rule. Differentite both the numerator and the denominator until you can substitute x = 0. The answer to the differentiated limit will be the answer to the original question.

lim (x -> 0) of (sin2x/x) = lim (x -> 0) (2cos2x/1) = 2*cos(2*0) = 2*cos0 = 2

Good luck!

=)

Answer 8

I'm not sure if this is what you're looking for, but here goes.

Using L'Hopital's Theorem : the limit as x tends to zero both sinx and x tends to ( goes to) zero.Therefore it is in a 0/0 form and by differntiation a simpler term can be obtained.

Differentiating sinx/x you will get 2cos2x/1
Which is basically 2cos2x.
As x tends to zero 2cos2x becomes 2 since cos(0)=1.

Answer 9

sinx = x - x^3/3! + x^5/5! - x^7/7! +....

For sin2x, this expansion is

sin2x = 2x - 8(x^3/3!) +....

sin2x/x has only one term free of x i.e. the first = 2
all other terms have x in them

so Limit as x approaches zero, sin2x / x = 2

Answer 10

A very easy way is to use the sine series:

sin x = x - x^3/3! + x^5/5!

So sin 2x = 2x - 8x^3/3! + ... (1)

Divide (1) by x:

sin 2x = 2 - 8x^2/3! + ...

As x approaches 0, sin 2x approaches 2.

Or you can use L'Hopitals rule.