Limit as x=0 of ((x+1)/2x)*sin(x)?

Question

Could you please do some algebraic manipulation to find the answer

Answer 1

The limit of ((x+1)/2x)*sin(x) as x -> 0 is as follows:


lim ((x+1) / (2*x)) * sin (x)
x->0

Using the distributive property, distribute the sin(x)

lim ((x * sin(x))/(2*x)) + (sin(x)/(2*x))
x->0


Now, we know that the limit of (sin(x) / x) = 1 as x ->0

So, factor out the (1/2) out of the two equations and we get:


lim (1/2)*[ ((x * sin(x)) / x) + (sin(x)/x) ]
x->0

The x's cancel out and we get

lim (1/2)*[ sin(x) + (sin(x)/x) ]
x->0

so we fill in the blanks,

lim (1/2) * [ sin(0) + (1) ], remember (sin(x) / x) = 1 as x ->0
x->0

lim (1/2) * (0 + 1) = (1/2)
x->0

The answer is 0.5 OR (1/2).

Hope this Helps.

Answer 2

Sin lim of x+1 as x->0 is 1, then in the limit, the function approaches, sin(x)/2x. Since lim sin(x)/x as x->0=1 (Surprise!), the limit of the function should approach 1/2.

Answer 3

this is ultimate proved by the Squeeze regulation. because -one million <= sin x <= one million for all x, we've that -one million/x <= sin x / x <= one million/x for all x removed from 0. because lim(x--> infinity) (+ or -) one million/x = 0, the Squeeze regulation facilitates us to end definitively that lim(x--> infinity) sin x / x = 0, besides. i'm hoping that helps!

Answer 4

the limit is as x approaches 0 is .5

If you make a table of values (x,y) and plug in numbers really close to 0 ( such as .000001, or -.000001) you can observe that x-value approaches .5