Question

lim x->0 4 * sinxcosx-sinx/x^2

Answer 1

\[\lim_{x \rightarrow 0}\]\[4\frac{ sinxcosx-sinx }{ x^2 }\]

Answer 2

If I just plug in, I get zero :(

Answer 3

you get 0/0 if you plug in 0

Answer 4

This is an indeterminant form that requires you to use LH rule.

Answer 5

What is LH rule? I haven't learn that yet.

Answer 6

\[4\lim_{s \rightarrow 0}\frac{\sin(x)\cos(x) - \sin(x)}{x^2}\]

Answer 7

\(\large\color{black}{\displaystyle\lim_{x \rightarrow ~0} \frac{\sin x \cos x- \sin x}{x^2}}\)

\(\large\color{black}{\displaystyle\lim_{x \rightarrow ~0} \frac{\sin x(\cos x- 1)}{x^2}}\)

\(\large\color{black}{\displaystyle\lim_{x \rightarrow ~0} \frac{\sin x}{x} \times \displaystyle\lim_{x \rightarrow ~0} \frac{\cos x- 1}{x}}\)

Answer 8

Good :)

Answer 9

so seems as though that it would still be zero:)
http://www.wolframalpha.com/input/?i=lim%5Bx-%3E+0%5D+%28sinx+%28cos+x+-+1%29%29%2Fx%5E2

Answer 10

and 0 times 4 is stiill zero:)

Answer 11

I am referring to the rules:

~ \(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin(x)}{x}=1}\)

~ \(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~0}\frac{1-\cos(x)}{x}=0}\)

Answer 12

your second limit, in black, is same as the rule number 2 in blue, but except that it is multiplied times -1, but we all know 0 times -1 is still a zero.

Answer 13

@Fanduekisses questions?

Answer 14

Ok so you plugged in, the cos(0) =1 and used the rules

Answer 15

no need plugging anything... I separated the limits, using a rule:
`A limit of a product is a product of limits.`

then I applied the 2 rules in blue

Answer 16

Where/how did you get the cosx-1 then?

Answer 17

they're just trig identities when you take the limit as your variable approaches 0

Answer 18

You factor our the \(\sin(x)\).

Answer 19

this site is shi(t) disconnects every time

Answer 20

I will post overview if you want

Answer 21

it was sinxcosx-sinx/x^2 then sinxcosx-1/x^2

Answer 22

Not exactly.

Answer 23

oooooooooohhhhhhhh

Answer 24

the sinx/sinx = 1

Answer 25

Only when x approaches 0.

Answer 26

By The Squeeze Theorem.

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @SolomonZelman
I am referring to the rules:

~ \(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin(x)}{x}=1}\)

~ \(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~0}\frac{1-\cos(x)}{x}=0}\)
\(\color{blue}{\text{End of Quote}}\)

Memorize this.

Answer 28

\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x\cos x-\sin x}{x^2}}\) factoring the top:


\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x(\cos x-1)}{x^2}}\)


re-writing as: \(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x}{x} \times \frac{\cos x-1}{x}}\)

limit separation, \(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x}{x} \times \displaystyle\lim_{x \rightarrow ~0}\frac{\cos x-1}{x}}\)


\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x}{x} \times \displaystyle\lim_{x \rightarrow ~0}\frac{-(1-\cos x)}{x}}\)


\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin x}{x} \times (-\displaystyle\lim_{x \rightarrow ~0}\frac{1-\cos x}{x})}\)

Answer 29

is all I did

Answer 30

then apply the blue rules

Answer 31

makes sense?

Answer 32

\(0 \cdot -1 = 0\) Typo.

Answer 33

-0 is still a zero 0... is more like it. but it's all same

Answer 34

Technically with limits, it means yuo are approaching 0 from the left.

Answer 35

your notation si wrong

Answer 36

don't write two math signs like this: separate with parenthesis:P
\(\large\color{black}{ 0 \cdot(-1)=0 }\) like this.

Answer 37

anyways, that ain't matters

Answer 38

the poster went offline, so apparently she got the needed assistance with the question.
(wow, I didn't disconnect just now, so unusual !!)