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 limit:limit [2014/12/12 21:21]jaan Fixed error: these formulas only hold for a > 1 limit:limit [2014/12/12 21:37] (current)jaan [Some properties of limit] Both sides previous revision Previous revision 2014/12/12 21:37 jaan [Some properties of limit] 2014/12/12 21:21 jaan Fixed error: these formulas only hold for a > 12014/01/13 21:02 marje created 2014/12/12 21:37 jaan [Some properties of limit] 2014/12/12 21:21 jaan Fixed error: these formulas only hold for a > 12014/01/13 21:02 marje created Line 32: Line 32: ===== Some properties of limit ===== ===== Some properties of limit ===== + ==== Limits of sum, difference, product and quotient ==== If $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ exist and are finite, then If $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ exist and are finite, then * $\lim_{x \to \infty} f(x) + g(x) = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)$, * $\lim_{x \to \infty} f(x) + g(x) = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)$, Line 39: Line 39: * $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\lim_{x \to \infty} f(x)}{\lim_{x \to \infty} g(x)}$ if $\lim_{x \to \infty} g(x) \neq 0$. * $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\lim_{x \to \infty} f(x)}{\lim_{x \to \infty} g(x)}$ if $\lim_{x \to \infty} g(x) \neq 0$. - For example, to calculate ​$\lim_{x \to \infty} \frac{8x^2 + x}{2x^2 - 1}$ we can express $\lim_{x \to \infty} ​\frac{8x^2 + x}{2x^2 - 1} = \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}}$ and then apply the rules above: $\lim_{x \to \infty} \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \frac{1}{x}}{\lim_{x \to \infty} 2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \lim_{x \to \infty} \frac{1}{x}}{\lim_{x \to \infty} 2 - \lim_{x \to \infty} \frac{1}{x^2}} = \frac{8 + 0}{2 - 0} = 4$. + + Compute ​$\lim_{x \to \infty} \frac{8x^2 + x}{2x^2 - 1}$. ++Solution| + We can express $\frac{8x^2 + x}{2x^2 - 1} = \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}}$ and then apply the rules above: $\lim_{x \to \infty} \frac{8x^2 + x}{2x^2 - 1} = \lim_{x \to \infty} \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \frac{1}{x}}{\lim_{x \to \infty} 2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \lim_{x \to \infty} \frac{1}{x}}{\lim_{x \to \infty} 2 - \lim_{x \to \infty} \frac{1}{x^2}} = \frac{8 + 0}{2 - 0} = 4$.++ + ​ When one of the limits is infinite, one can use the following rules: When one of the limits is infinite, one can use the following rules: Line 48: Line 51: * $q / \infty = 0$ if $q \neq \pm \infty$. * $q / \infty = 0$ if $q \neq \pm \infty$. - For example, ​$\lim_{x \to \infty} 5 + x = \lim_{x \to \infty} 5 + \lim_{x \to \infty} x = 5 + \infty = \infty$. + ​Compute $\lim_{x \to \infty} 5 + x$. ++Solution| + $\lim_{x \to \infty} 5 + x = \lim_{x \to \infty} 5 + \lim_{x \to \infty} x = 5 + \infty = \infty$.++ + ​ In other cases, like In other cases, like Line 58: Line 63: If $\lim_{x \to \infty} g(x) = 0$ then, as said above, knowing only $\lim_{x \to \infty} f(x)$ is not enough to tell what $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ is. However, the situation is better if we also know the signs of $f(x)$ and $g(x)$. For example, $\lim \frac{x}{e^{-x}} = \frac{\infty}{+0} = \infty$, $\lim \frac{x}{-e^{-x}} = \frac{\infty}{-0} = -\infty$. Here $+0$ and $-0$ denote the facts that $\lim_{x \to \infty} e^{-x} = \lim_{x \to \infty} -e^{-x} = 0$ but $e^{-x}$ converges to $0$ from the positive side (i. e. $e^{-x} > 0$) and $-e^{-x}$ converges to $0$ from the negative side. If $\lim_{x \to \infty} g(x) = 0$ then, as said above, knowing only $\lim_{x \to \infty} f(x)$ is not enough to tell what $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ is. However, the situation is better if we also know the signs of $f(x)$ and $g(x)$. For example, $\lim \frac{x}{e^{-x}} = \frac{\infty}{+0} = \infty$, $\lim \frac{x}{-e^{-x}} = \frac{\infty}{-0} = -\infty$. Here $+0$ and $-0$ denote the facts that $\lim_{x \to \infty} e^{-x} = \lim_{x \to \infty} -e^{-x} = 0$ but $e^{-x}$ converges to $0$ from the positive side (i. e. $e^{-x} > 0$) and $-e^{-x}$ converges to $0$ from the negative side. - Some well-known limits: + ==== Some well-known limits ​==== * $\lim_{x \to \infty} \frac{x^d}{a^{x}} = 0$ for any constants $a > 1, d > 0$. * $\lim_{x \to \infty} \frac{x^d}{a^{x}} = 0$ for any constants $a > 1, d > 0$. * $\lim_{x \to \infty} \frac{x^d}{(\log_a x)^b} = \infty$ for any constants $a > 1, b > 0, d > 0$. * $\lim_{x \to \infty} \frac{x^d}{(\log_a x)^b} = \infty$ for any constants $a > 1, b > 0, d > 0$.