Ok so this is probably the easiest thing ever, but I've completely blanked on how to do this:
lim x (approaches) 0 for sinx over x
I was just substituting 0 in for x but I checked the answers page and it said it equals 1???
Thanks ♥
lim x (approaches) 0 for sinx over x
I was just substituting 0 in for x but I checked the answers page and it said it equals 1???
Thanks ♥
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If you substitute in x = 0, you will get sin(0)/0 = 0/0, which is indeterminate and, hence, you cannot find the value of the limit without further investigation.
lim (x-->0) sin(x)/x = 1 is actually a special trigonometric limit. It's proof is usually presented alongside the Squeeze Theorem (which is what is used to prove this limit). Basically, the proof is outlined like this: you want to find two functions that go to 1 as x --> 0: one that is less than sin(x)/x and one that is more than sin(x)/x. Using a unit circle and by drawing a few triangles, you can use the areas to conclude that:
(1/2)sin(x) < x/2 < (1/2)tan(x), for 0 < x < π/2
==> cos(x) < sin(x)/x < 1.
Since lim (x-->0+) cos(x) = 1 and lim (x-->0+) 1 = 1, the Squeeze Theorem tells us that:
lim (x-->0+) sin(x)/x = 1.
Using the fact that sin(x)/x is odd, lim (x-->0-) sin(x)/x = 1 follows. Therefore:
lim (x-->0) sin(x)/x = lim (x-->0+) sin(x)/x = lim (x-->0-) sin(x)/x = 1. Q.E.D.
I hope this helps!
lim (x-->0) sin(x)/x = 1 is actually a special trigonometric limit. It's proof is usually presented alongside the Squeeze Theorem (which is what is used to prove this limit). Basically, the proof is outlined like this: you want to find two functions that go to 1 as x --> 0: one that is less than sin(x)/x and one that is more than sin(x)/x. Using a unit circle and by drawing a few triangles, you can use the areas to conclude that:
(1/2)sin(x) < x/2 < (1/2)tan(x), for 0 < x < π/2
==> cos(x) < sin(x)/x < 1.
Since lim (x-->0+) cos(x) = 1 and lim (x-->0+) 1 = 1, the Squeeze Theorem tells us that:
lim (x-->0+) sin(x)/x = 1.
Using the fact that sin(x)/x is odd, lim (x-->0-) sin(x)/x = 1 follows. Therefore:
lim (x-->0) sin(x)/x = lim (x-->0+) sin(x)/x = lim (x-->0-) sin(x)/x = 1. Q.E.D.
I hope this helps!
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It is actually a rule to remember!!
ie. lim (x->0) (sin x)/x = 1
The proof would have been given to you in class. It is too complicated to do in full here, but it stems from the fact that (provided x is in radians) that sin x is approximately equal to x when x is small.
ie. lim (x->0) (sin x)/x = 1
The proof would have been given to you in class. It is too complicated to do in full here, but it stems from the fact that (provided x is in radians) that sin x is approximately equal to x when x is small.