Remarkable Limits

A techinique for calculation of limits consists in restoring to limits already knwon said remarkable limits. Tale strategia permette di risolvere ampie gamme di limiti che presentano forme di indecizione.

Proposition 4.10.1 The following holds

lim_{x \to 0} \frac{\sin x}{x} = 1

Proof. It is an indeterminate form 0/0. We use Theorem 4.1.8 to show that the right-hand and left-hand limits are both 1. Observe that sin x and x are both odd functions. Therefore f(x) = (sin x) / x is an even function, with a graph symmetric about the y-axis. This symmetry implies that the left-hand limit at 0 if exits is equal to the right-hand limit.

lim_{x \to 0^{-}} \frac{\sin x}{x} = lim_{x \to 0^{+}} \frac{\sin (- x)}{- x} = lim_{x \to 0^{+}} \frac{- \sin x}{- x} = lim_{x \to 0^{+}} \frac{\sin x}{x}

we shall focus on the right-hand limit. Take a look at the followin diagram:

The are of the triangle OCA is less than that of the circular sector OCA which in turn is less than that of the triangle OBA. We can express the areas as

Area Δ OPA = 1/2 ⋅ base ⋅ height = 1/2 ⋅ 1 ⋅ sin x
Area sector OCA (radius r) = 1/2 ⋅ r² x = x/2
Area Δ OBA = 1/2 ⋅ base ⋅ height = 1/2 ⋅ 1 ⋅ tg x

It follows that:

sin x/2 < x/2 < tg x/2

that is

sin x < x< tg x

dividing by sin x, we have:

1 < x / sin x < 1/ cos x

Taking reciprocals reverses the inequalities:

1 > sin x/x > cos x

Since lim_{x ⟶ 0⁺} cos x = 1, the sandwich theorem gives

lim_{x \to 0^{+}} \frac{\sin x}{x} = 1

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Proposition 4.10.2 We have that

lim_{x \to 0} \frac{\tan x}{x} = 1

Proof. From the definition of tangent

lim_{x \to 0} \frac{\tan x}{x} = lim_{x \to 0} \frac{\sin x}{\cos x} \cdot \frac{1}{x} = lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = lim_{x \to 0} \frac{1}{\cos x} = 1 □

Proposition 4.10.3 We have that

lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}

Proof. We have that

\frac{1 - \cos x}{x^{2}} = \frac{1 - \cos^{2} x}{x^{2} (1 + \cos x)} = {(\frac{\sin x}{x})}^{2} \frac{1}{1 + \cos x} \to \frac{1}{2} □

Definition 4.10.4 (Asymptotic comparison) We say that f ~ g as x ⟶ x₀ if the limit of the ratio of the two function is 1, that is

lim_{x \to x_{0}} \frac{f (x)}{g (x)} = 1

The limits we have studied can be written with this notation as

sin x ~ x 1 − cos x ~ x²/2 e^x − 1 ~ x log(1 + x) ~ x

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