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Lim x- 0 sinx/x proof-Lim (x^2 2x 3)/(x^2 2x 3) as x > 3;1) ∞ 2) 1 3) 0 4) does not exist Solution (3) 0 \(\begin{array}{l} \lim _{x \rightarrow \infty} \frac{\sin x}{x}\\ \sin \infty \in1,1 \text { oscillatory

Prove Lim Sin X X 1 As X Approaches 0 Squeeze Theorem Youtube

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Sin(x) lim = 1 x→0 x In order to compute specific formulas for the derivatives of sin(x) and cos(x), we needed to understand the behavior of sin(x)/x near x = 0 (property B) In his lecture, Professor Jerison uses the definition of sin(θ) as the ycoordinate of a point on the unit circle to prove that lim θ→0(sin(θ)/θ) = 1For example, lim x→0 2x cosx = 0 1 = 0, but a blind application of L'Hˆopital's rule leads incorrectly to lim x→0 2x cosx To solve it, we can apply the L'Hôpital's rule Given two functions f and g differentiable at x = a, it holds that If lim x→a f (x) g(x) = 0 0 or lim x→a f (x) g(x) = ∞ ∞ then lim x→a f (x) g(x) = lim x→a f '(x) g'(x) So we have lim x→0 x

The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0 If we wish to calculate the Taylor series at any other value of x , we can consider a variety of approaches Suppose we wish to find the Taylor series of sin( x ) at x = c , where c is any real number that is not zeroLim x!0 sin(x) x = 1 using an proof Solution One can see that the following inequalities are true for values close to zero, both positive and negative jsin(x)jLim 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 2

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 Requirement lim x→0 sinx x = 1 lim x → 0 sin ⁡ x x = 1 Proof of limit of sin x / x = 1 as x approaches 0 Proof By definition of tan x tanx x = sinx x ⋅cosx = sinx x × 1 cosx tan ⁡Cosx < sinx / x < 1 Hence proved Using the above theorem we can easily prove some other trigonometric identities such as lim x>0 sinx/x =1;

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