 # Limits of Functions

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} = -∞\$

## some important limits

\${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\$

## Asymptotes 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)\$

### Example 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.

### Example 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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