Section 2.3

Theorem: If f(x) ≤ g(x) near a (except possibly at a) and if lim_{x→a}f(x) and lim_{x→a}g(x) exist, then

lim_{x→a}f(x) ≤ lim_{x→a}g(x)

Squeeze theorem: If f(x) ≤ g(x) ≤ h(x) near a (except possibly at a) and if lim_{x→a}f(x) = L and lim_{x→a}h(x) = L, then

lim_{x→a}g(x) = L

Example: g(x) = xsin\frac{1}{x}

−|x| ≤ xsin\frac{1}{x} ≤ |x|
lim_{x→0}( − |x|) = 0, lim_{x→0}(|x|) = 0.

Hence , lim_{x→0}(xsin\frac{1}{x}) = 0 Section 2.4 Informal Defn: lim_{x→a}f(x) = L if x close to a (except possibly at a) implies f(x) is close to L.

Formal Defn: lim_{x→a}f(x) = L if For all ϵ > 0, there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ϵ.

Formal Defn: lim_{x→a}f(x) = L if For all ϵ > 0, there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ϵ. Proof: Let ϵ > 0. Choose δ = . Note δ = > 0. Suppose 0 < |x − a| < δ. Claim: |f(x) − L| < ϵ.

Defn: lim_{x→a}f(x) = L if for all ϵ > 0, there exists a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ϵ

Show lim_{x→1}2 = Show lim_{x→4}2x + 3 = Defn: lim_{x→a^{−}}f(x) = L if x close to a and x < a implies f(x) is close to L.

Defn: lim_{x→a^{+}}f(x) = L if x close to a and x > a implies f(x) is close to L.

Defn: lim_{x→a}f(x) = ∞ if x close to a (except possibly at a) implies f(x) is large.

Defn: lim_{x→a}f(x) = −∞ if x close to a (except possibly at a) implies f(x) is negative and |f(x)| is large.

Section 2.5 Defn: f is continuous at a if lim_{x→a}f(x) = f(a) (i.e., if lim_{x→a}f(x) = f(lim_{x→a}x) Examples:

Ex: Polynomial, rational, root, trigonometric, inverse trigonometric, exponential, logarithmic functions are continuous functions. Read left and right continuity If f, g continuous at a, c ∈ R, then f + g, fg, cf, f/g (if g(a) ≠ 0) are continuous.

If g continuous at a and f continuous at g(a), then f ∘ g continuous at a.

Ex: lim_{x→0} \frac{x^{2}−e^{x^{3}}}{cos(x)} =

Intermediate value theorem: Suppose f continuous on [a, b], f(a) ≠ f(b) and n is between f(a) and f(b), then there exists c ∈ (a, b) such that f(c) = N.

Example: Show that x^{2} − 7x + 1 has a root between 0 and 1.

Section 2.3: To find vertical asymptotes, find all a ∈ R such that lim_{x→a^{−}}f(x) = ±∞ and/or lim_{x→a^{+}}f(x) = ±∞

Ex: f(x) = \frac{1}{(x+2)(x−3)^{2}}

Section 2.6: Horizontal asymptotes/limits at infinity

To find horizontal asymptotes: calculate lim_{x→+∞}f(x) and lim_{x→−∞}f(x)

IF lim_{x→+∞}f(x) = L where L is a finite real number, then y = L is a horizontal asymptote.

IF lim_{x→−∞}f(x) = K where K is a finite real number, then y = K is a horizontal asymptote.

Ex: f(x) = \frac{2x^{3}−x^{2}+1}{8x^{3}+x+3} lim_{x→+∞} \frac{2x^{3}−x^{2}+1}{8x^{3}+x+3} =

Similarly, lim_{x→−∞} \frac{2x^{3}−x^{2}+1}{8x^{3}+x+3} = Horizontal asymptote(s):

Ex: f(x) = \frac{x^{2}+1}{2x^{5}+x^{2}−3} lim_{x→+∞} \frac{x^{2}+1}{2x^{5}+x^{2}−3} =

Similarly, lim_{x→−∞} \frac{x^{2}+1}{2x^{5}+x^{2}−3} = Horizontal asymptote(s): Ex: f(x) = \frac{2x^{5}+x^{2}−3}{x^{2}+1} lim_{x→+∞} \frac{2x^{5}+x^{2}−3}{x^{2}+1} =

Also, lim_{x→−∞} \frac{2x^{5}+x^{2}−3}{x^{2}+1} = Horizontal asymptote(s):

Ex: f(x) = \frac{2x}{\sqrt{x^{2}+1}} lim_{x→+∞} \frac{2x}{\sqrt{x^{2}+1}} =

lim_{x→−∞} \frac{2x}{\sqrt{x^{2}+1}} =

Horizontal asymptote(s):

Ex: f(x) = x^{2} − x^{3} lim_{x→+∞}x^{2} − x^{3} = lim_{x→−∞}x^{2} − x^{3} = Horizontal asymptote(s): Ex: f(x) = x^{\frac{2}{3}} − x lim_{x→+∞}x^{\frac{2}{3}} − x = lim_{x→−∞}x^{\frac{2}{3}} − x = Horizontal asymptote(s):