Question

Find the limit (No L'hopital rule): written down.

limx→0sinx1−2sin(x+π6)

Answer 1

0/0

Answer 2

Bro, I tried all this.

Answer 3

lol!!

i will have a go too, interesting

Answer 4

sin(x+30) = sinxcos30 + sin30cosx
1-2sin(x+30) = 1- sqrt(3)sinx + cosx
sinx/ 1- sqrt(3)sinx + cosx

Answer 5

Give it a go, and tell me what was the result.

Answer 6

as it stands, im finding it difficult to interpet your function
can you use latex and put it in more friendly form ?

Answer 7

\[\lim_{x \rightarrow 0} \frac{ sinx }{ 1-2\sin(x+\frac{ \pi }{ 6 }) }\]

Answer 8

break it up as you did before (but correct the sin on the bottom)

ie it should be
sinx/ 1- sqrt(3)sinx - cosx

and limit should be -1/ root(3)

Answer 9

right it's -1/sqrt(3)
what did you do ?

Answer 10

exactly what you did. break it up and work through.
you get to x/-root(3)x, right?
lim x/x = 1

Answer 11

I can't.

Answer 12

@IrishBoy123

Answer 13

\[\frac{sinx}{1- 2\sin(x) \cos \frac{π}{6} - 2\cos(x) \sin \frac{π}{6}}\]
\[\frac{sinx}{1 - 2 \sin(x) \frac {\sqrt 3}{2} - 2 \cos(x) \frac{1}{2}}\]
\[\frac{x}{1 - \sqrt{3} x - 1}\]

Answer 14

sinx1−2sin(x)3√2−2cos(x)12

sinx/ 1- sqrt(3)sinx - cosx
last step ?

Answer 15

@IrishBoy123 , What kind of magic did you do to get x/1-sqrt3)x - 1

Answer 16

sin x ->x, cos x ->1 as x ->0

Answer 17

Right the steps.

Answer 18

those are the steps, there is little else too it apart maybe from finding the limit symbol in the equation drawing thingy, which i have never been able to do and stuffing that on the LHS of the last one before introducing the limits.


here are some more:
http://planetmath.org/listofcommonlimits

Answer 19

actually let me try

\[\lim_{x \rightarrow 0} \frac{sinx}{x} = 1\]

et voila!

Answer 20

sinx/x / 1/x - sinx/x * sqrt(3) - cosx/x
now ???
1/ 1/0 - 1/0
undefined - undefined

Answer 21

OK

\[\lim_{x \rightarrow 0} \frac{sinx}{1 - \sqrt{3}sinx - \cos x} = \frac{x}{1 - \sqrt{3}x - 1} = -\frac{x}{\sqrt{3}x} = - \frac{1}{\sqrt{3}}\]

Answer 22

Bro, as soon as you apply that sinx/x =1 you must put each x =0. You can't multiply by x.

Answer 23

these are limits!

Answer 24

The final result is right, but the steps are defected.

Answer 25

the limit of a quotient is the quotient of the limits
in other words, you can do this
\[ \lim_{x \rightarrow 0} \frac{\sin x}{1 - \sqrt{3}\sin x - \cos x} \\
\frac{ \lim_{x \rightarrow 0} \sin x} {\lim_{x \rightarrow 0}( 1 - \sqrt{3}\sin x - \cos x)}
\]
also, the limit of a sum is the sum of the limits

Answer 26

@TrojanPoem


"The final result is right, but the steps are defected. "

you are the re-incarnation of Bishop Berkeley?!?! right ?!?!

Answer 27

@IrishBoy123 , Excuse me, thanks.
@phi, Thanks , I though I can't do that.

Answer 28

When you have time, see
http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-5-a-more-rigorous-approach-to-limits/

(you might want to also watch the lecture previous to this one, which provides the background and lots of intuition)

Answer 29

Thanks that's what I needed.

Answer 30

to figure out the limit without using L'Hopital, we could use series expansions
once you get to
\[ \lim_{x \rightarrow 0} \frac{\sin x}{1 - \sqrt{3}\sin x - \cos x} \]
I would rewrite it as
\[ \lim_{x \rightarrow 0} \frac{\frac{\sin x}{x}}{\frac{1 - \cos x}{x} - \sqrt{3}\frac{\sin x}{x}} \]

and now take the limit of the top and bottom
the limit of the numerator is
\[ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\]

the limit of the bottom is
\[ \lim_{x \rightarrow 0}\frac{1 - \cos x}{x} -\sqrt{3}\lim_{x \rightarrow 0}\frac{\sin x}{x} \]

as before the second term has a limit of -sqr(3)*1

for the first term, use
\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...\]
so
\[ 1 - \cos x= \frac{x^2}{2!} - \frac{x^4}{4!}+ \frac{x^6}{6!} - ...\]
and
\[ \frac{1- \cos x}{x} = \frac{x}{2!} - \frac{x^3}{4!}+ \frac{x^5}{6!} - ...\]

now take the limit
\[ \lim_{x \rightarrow 0}\frac{1 - \cos x}{x} =\lim_{x \rightarrow 0} \left( \frac{x}{2!} - \frac{x^3}{4!}+ \frac{x^5}{6!} -...\right) =0 \]

putting it all together
\[ \lim_{x \rightarrow 0} \frac{\frac{\sin x}{x}}{\frac{1 - \cos x}{x} - \sqrt{3}\frac{\sin x}{x}} = \frac{1}{0 - \sqrt{3} \cdot 1}= - \frac{\sqrt{3}}{3}\]

Answer 31

another way to find the limit of
\[ \lim_{x \rightarrow 0}\frac{1 - \cos x}{x} \\
= \lim_{x \rightarrow 0}\frac{1 - \cos x}{x} \frac{(1+\cos x)}{(1+\cos x)} \\
= \lim_{x \rightarrow 0}\frac{1 - \cos^2 x}{x(1+\cos x)} \\
= \lim_{x \rightarrow 0}\frac{\sin x}{x} \frac{\sin x}{(1+\cos x)}
\]
and now we can do
\[ \lim_{x \rightarrow 0}\frac{\sin x}{x} \lim_{x \rightarrow 0}\frac{\sin x}{(1+\cos x)} \]
to get 1 * 0 = 0

Answer 32

Thanks, I didn't take the first one which involve 4!