Proof of lim sinx/x Hi all, math noob here Where all limits are lim x>0 So lim sinx/x = 1 5 Share Report Save level 1 5y You'll wanna check out the Squeeze Theorem which is incredibly useful in many situations, and applies directly to your question here ) 1 ShareSqueeze Theorem Suppose that g(x)≤f(x)≤h(x) g ( x) ≤ f ( x) ≤ h ( x) for all x x close to a a but not equal to a a If lim x→ag(x)= L= lim x→ah(x), lim x → a g ( x) = L = lim x → a h ( x), then lim x→af(x)= L lim x → a f ( x) = L This theorem can be proved using the official definition of limit− O ( x 7) x = lim x → 0 ( 1 − x 2 3!
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Lim x- 0 sin(x)/x = 1 proof
Lim x- 0 sin(x)/x = 1 proof-How do you prove lim x tends to 0 sinx/x using expansion? lim (θ → 0) sin θ / θ = 1 Proof Consider a circle with centre 'O' and radius 'r' Mark two point A and B on the circumference of the circle so that (AOB)!
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⋯) sin x x = 1 − x 2 3!Popular Problems Calculus Evaluate limit as x approaches 0 of (sin (x^2))/x lim x→0 sin (x2) x lim x → 0 sin ( x 2) x Evaluate the limit of the numerator and the limit of the denominator Tap for more steps Take the limit of the numerator and the limit of the denominatorMotivation for the lim sin(x)/x as x to 0 Why sin(x)/x tends to 1 The following short note has appeared in a 1943 issue of the American Mathematical Monthly
Lim x→0 sin(x) x =1 To do this, we'll use the Squeeze theorem by establishing upper and lower bounds on sin(x)~x in an interval around 0 Speci cally, we'll show that cos(x) ≤ sin(x) x ≤1 in an interval around 0 We can already see why this should be the case by the following graph y=cos(x) y= sin(x) x y=1 2This mainly uses the special technique that I discovered while writing this There are a few things that we need (1) lim x → 0 sin x x = 1 (2) sin ( 3 x) = 3 sin x − 4 sin 3 x (3) tan ( 3 x) = 3 tan x − tan 3 x 1 − tan 2 x We will also need (4) lim x → 0 tan x x = 1= lim x → 0 x 2 sin 1/ x = (lim x → 0 x) (lim x → 0 x sin 1/ x) = (0) (0) (using the fact that f is continuous at 0) = 0 So g is differentiable at 0, with derivative 0 Definition We say that a function f is differentiable on the open interval I if f is differentiable at x for every x in I
At the start of the lecture we saw an algebraic proof that the derivative of sin x is cos x While this proof was perfectly valid, it was somewhat abstract – it did not make use of the definition of the sine function sin θ The proof that lim = 1 did use the unit circle definition of the sine of θ→0Lim (sin x x)/x^3 as x>0 Natural Language;Lim x>0 (1 – cosx)/x =0;
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x x ≤ 1, ∀ x ∈ − π 2, 0 ∪ 0, π 2 By using the Squeeze Theorem lim x→0 sinx x = lim x→0cosx = lim x→01 = 1 lim x → 0 sin x x = lim x → 0 cos x = lim x → 0 1 = 1 we conclude that lim x→0 sinx x = 1 lim x → 0 sinA c c o r d i n g t o T a y l o r (o r) M a c l a u r i n E x p a n s i o n, s i n e f u n c t i o n c a n b e g i v e n a s sin x = x − x 3 3!Taking the limit now we get it as 1 Btw we were not taught this just that they asked us to memorize this result I just thought of this proof as I already did Taylor Series in SHM Chapter in Physics
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Prove that \lim_{x\rightarrow} \frac{\sin x}{x} = 1 Solution Given \epsilon > 0 want to find \delta such that \left\frac{\sin x}{x} 1 \right= θ At 'A' draw a tangent to the circle produce OB to cut the tangent at CLimit as x approaching 0 of (sin (x))/x \square!
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This is because the value of (sinx/x),when x approaches to zero is very slightly lesser than 1 Thus when you apply greatest integer function to3 Problem 3 Show that lim x!0 sin(1=x) does not exists, using an proof Solution The easiest way is a proof by contradiction Suppose the limit did exist, then there would be an Lsuch that given an >0, then jxj< would implyIt is not clear what the limit is In fact, depending on what functions f ( x) and g ( x) are, the limit
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I knew that if I show that each limit was 1, then the entire limit was 1 I decided to start with the lefthand limit For x− x 6 7O ( x 4)) = 1 There are even
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