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limit:limit [2014/01/13 21:02] marje created |
limit:limit [2014/12/12 21:37] jaan [Some properties of limit] |
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| ===== 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 | ||
| * $\lim_{x \to \infty} f(x) + g(x) = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)$, | * $\lim_{x \to \infty} f(x) + g(x) = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)$, | ||
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| * $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\lim_{x \to \infty} f(x)}{\lim_{x \to \infty} g(x)}$ 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: $\lim_{x \to \infty} \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \frac{1}{x}}{\lim_{x \to \infty} 2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \lim_{x \to \infty} \frac{1}{x}}{\lim_{x \to \infty} 2 - \lim_{x \to \infty} \frac{1}{x^2}} = \frac{8 + 0}{2 - 0} = 4$. | + | <WRAP task> |
| + | 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: $\lim_{x \to \infty} \frac{8x^2 + x}{2x^2 - 1} = \lim_{x \to \infty} \frac{8 + \frac{1}{x}}{2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \frac{1}{x}}{\lim_{x \to \infty} 2 - \frac{1}{x^2}} = \frac{\lim_{x \to \infty} 8 + \lim_{x \to \infty} \frac{1}{x}}{\lim_{x \to \infty} 2 - \lim_{x \to \infty} \frac{1}{x^2}} = \frac{8 + 0}{2 - 0} = 4$.++ | ||
| + | </WRAP> | ||
| 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: | ||
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| * $q / \infty = 0$ if $q \neq \pm \infty$. | * $q / \infty = 0$ if $q \neq \pm \infty$. | ||
| - | For example, $\lim_{x \to \infty} 5 + x = \lim_{x \to \infty} 5 + \lim_{x \to \infty} x = 5 + \infty = \infty$. | + | <WRAP task>Compute $\lim_{x \to \infty} 5 + x$. ++Solution| |
| + | $\lim_{x \to \infty} 5 + x = \lim_{x \to \infty} 5 + \lim_{x \to \infty} x = 5 + \infty = \infty$.++ | ||
| + | </WRAP> | ||
| In other cases, like | In other cases, like | ||
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| 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 $a > 0, d > 0$. | + | * $\lim_{x \to \infty} \frac{x^d}{a^{x}} = 0$ for any constants $a > 1, d > 0$. |
| - | * $\lim_{x \to \infty} \frac{x^d}{(\log_a x)^b} = \infty$ for any constants $a > 0, b > 0, d > 0$. | + | * $\lim_{x \to \infty} \frac{x^d}{(\log_a x)^b} = \infty$ for any constants $a > 1, b > 0, d > 0$. |
| ===== Exercises ===== | ===== Exercises ===== | ||