limit of
sin x
as x approaches 0∗
x
Wkbj79†
2013-03-21 22:38:53
Theorem 1.
lim
x→0
sin x
=1
x
π
Note that this entry uses the result x < tan x for 0 < x < . I have
2
received word that a proof of this fact which uses no calculus will be supplied
on PlanetMath. Please let me know when this is completed so that I can edit
this entry accordingly.
Proof. First, let 0 < x <
π
. Then 0 < cos x < 1. Note also that
2
x < tan x.
(1)
Multiplying both sides of this inequality by cos x yields
x cos x < sin x.
(2)
sin x < x.
(3)
By this theorem,
Combining inequalities (??) and (??) gives
x cos x < sin x < x.
(4)
Dividing by x yields
sin x
< 1.
(5)
x
−π
π
Now let
< x < 0. Then 0 < −x < . Plugging −x into inequality (??)
2
2
gives
cos x <
∗ hLimitOfdisplaystylefracsinXxAsXApproaches0i
created: h2013-03-21i by: hWkbj79i
version: h39255i Privacy setting: h1i hTheoremi h26A06i h26A03i
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1
sin(−x)
< 1.
−x
Since cos is an even function and sin is an odd function, we have
cos(−x) <
cos x <
− sin x
< 1.
−x
(6)
(7)
π
Therefore, inequality (??) holds for all real x with 0 < |x| < .
2
Since cos is continuous, lim cos x = cos 0 = 1. Thus,
x→0
sin x
≤ lim 1 = 1.
x→0
x
sin x
= 1.
By the squeeze theorem, it follows that lim
x→0 x
1 = lim cos x ≤ lim
x→0
x→0
(8)
Note that the above limit is also valid if x is considered as a complex variable.
2