Given the polynomials:

$$p(x) = a_kx^k + a_{k-1}x^{k-1} + ... + a_1x + a_0$$ $$q(x) = b_mx^m + b_{m-1}x^{m-1} + ... + b_1x + b_0$$then:

$${lim}↙{x→x_0} p(x) = p(x_0) $$ $${lim}↙{x→x_0} {p(x)}/{q(x)} = {p(x_0)}/{q(x_0)}$$if $q(x_0) ≠ 0$$ $${lim}↙{x→±∞} {p(x)}/{q(x)} = [\table ± ∞, k > m; {a_k}/{b_m}, k = m; 0, k = m, k < m;$$

e.g. ${lim}↙{x→± ∞} {3x^2}/{x^2 + 5}$

In this case, k = 2 and m = 2, so ${lim}↙{x→± ∞} {3x^2}/{x^2 + 5} = {a_k}/{b_m} = {3}/{1} = 3$

e.g. ${lim}↙{x→5} {-x^2 + 5x}/{x^2 - 2x - 15} = {-x(x - 5)}/{(x - 5)(x + 3)} = {-x}/{(x + 3)} = -5/8$

${lim}↙{x→∞} {x^3 - 1}/{x^2 + x}$ | and, ${lim}↙{x→-∞} {x^3 - 1}/{x^2 + x}$ |

In this case, k = 3 and m = 2, so ${lim}↙{x→∞} {x^3 - 1}/{x^2 + x} = ∞$ | ${lim}↙{x→-∞} {x^3 - 1}/{x^2 + x} = -∞$ |

${lim}↙{x→0} {sinx}/x = 1$

${lim}↙{x→± ∞} (1 + 1/x)^x = e$

${lim}↙{x→0} {log_a(1 + x)}/x = log_ae$, provided $0 < a ≠ 1$

${lim}↙{x→0} {ln(1 + x)}/x = 1$

${lim}↙{x→0} {a^x - 1}/x = ln (a)$, provided $0 < a ≠ 1$

${lim}↙{x→0} {e^x - 1}/x = 1$

If ${lim}↙{x→∞} (f(x)-g(x)) = 0$, g approximates f asymptotically when $x→∞$

$m = {lim}↙{x→∞} {f(x)}/x$

$q = {lim}↙{x→∞} (f(x) - m⋅x)$

f: x|→ ${x^3 - 2x^2 - x + 3}/{x^2 - 1}$, $D_f$ = R\{±1}

Vertical asymptotes: x = ± 1

Oblique asymptote:

$m = {lim}↙{x→±∞} {f(x)}/x = {x^3 - 2x^2 - x + 3}/{x^3 - x} = 1$

$q = {lim}↙{x→±∞} (f(x) - m⋅x)$

$= {lim}↙{x→±∞} {{x^3 - 2x^2 - x + 3 - x^3 + x}/{x^2 - 1} = -2$

Therefore, y = x - 2 is the oblique asymptote.

g: x|→ ${x^5 - x + 1}/{x^3 + x}$, $D_g = R^+$

Vertical asymptote: x = 0

Carrying through the division:

$g(x) = x^2 - 1 + 1/{x^3 + x}$

For which ${lim}↙{x→±∞} (g(x) - x^2 + 1)$

$ = {lim}↙{x→±∞} 1/{x^3 + x} = 0$

The function is therefore approximately asymptotic to the parabola $y = x^2 - 1$

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