Question

Find the Limit:
lim x-0 (sinsxsin5x)/(x^2)

Answer 1

familiar with the limit :
\[\large \lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1\]

?

Answer 2

yes

Answer 3

use it

Answer 4

how? can I pull a sinx out of the top? making it sinx(15)/x^2? then i could make it (sinx/x)*(15/X)?

Answer 5

\[\large \lim\limits_{x\to 0} \dfrac{\sin 3x~\sin 5x}{x^2}\]

Answer 6

is that the limit you're trying to evaluate ?

Answer 7

yes!
I just realized I had a typo in the original

Answer 8

\[\large \begin{align}\lim\limits_{x\to 0} \dfrac{\sin 3x~\sin 5x}{x^2} &=15~\lim\limits_{x\to 0} \dfrac{\sin 3x}{3x}\dfrac{\sin 5x}{5x} \end{align} \]

Answer 9

split that into product of two limits
each evaluates to 1

Answer 10

\[\large \begin{align}\lim\limits_{x\to 0} \dfrac{\sin 3x~\sin 5x}{x^2} &=15~\lim\limits_{x\to 0} \dfrac{\sin 3x}{3x}\dfrac{\sin 5x}{5x} \\~\\&= 15~\left( \lim\limits_{3x\to 0} \dfrac{\sin 3x}{3x}\right) ~ \left(\lim\limits_{5x\to 0} \dfrac{\sin 5x}{5x} \right)\\~\\ &= 15~\left( 1\right) ~ \left(1\right)\\~\\&= 15\end{align} \]

Answer 11

I forgot you can split the limits like that and multiply a number through like that... thank you so much! I didn't even think about looking back at my limit rules.. =(

Answer 12

we cannot do that everytime, but we can do it whenever we know that both the limits exist

Answer 13

\[\large \lim\limits_{x\to a} f(x)\times g(x) = \lim\limits_{x\to a} f(x)\times \lim\limits_{x\to a} g(x) \]

is true only when both the limits on right hand side exist

Answer 14

it may not look like a big deal, but it can get us into weird problems if we are not careful..

Answer 15

so if the limit doesn't exist you cannot split them because it would be like multiplying by something that isn't there?

Answer 16

Exactly !

Answer 17

sorry I had to put that into simple english. Thank you so much! I know I am new but I love algebra and as a user am I allowed to go help others?

Answer 18

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Answer 19

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Answer 20

Thanks!