Question

find the limit

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin 2x }{ x }\]


We were taight the sandwhich theorem but I am not sure how to do it for this limit

Answer 2

taught*

Answer 3

1

Answer 4

I need to show algebraically

Answer 5

we were taught to set it up as an inequality so
\[-1\le \frac{ \sin2x }{ x }\le1 \]

Answer 6

and then multiply through by the equation you sandwich it in and it would work for \[x ^{2}+1\]\[x ^{2}-1\] but cant you multiply through by only one equation?

Answer 7

and wait it is 2 not 1

Answer 8

so replace the x^2+1 with x^2+2 and same with the negative

Answer 9

sorry it should be \[-x ^{2}-2\]

Answer 10

Ok I get the first thing but the next two what?

Answer 11

All you need is the angle of sin and something in the denominator to match. If you have sin(2x), then you need 2x in the denominator somehow. So how do you get a 2 in the denominator?

Answer 12

multiply throughout by 2 but doesnt that just leve me with sin?