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I'm looking for this type of indeterminations and what I have to do to solve these:

$\lim\limits_{x\to\infty}{f(x)=\dfrac{\infty}{\infty}}$

$\lim\limits_{x\to\infty}{f(x)=\infty-\infty}$

$\lim\limits_{x\to\infty}{f(x)=\dfrac{0}{0}}$

$\lim\limits_{x\to\infty}{f(x)=\dfrac{n}{0}},n\not=0$


EDIT: Could anyone post any examples of these indeterminations?

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Just examples? (1) $f(x) = \frac{x^2+1}{x^3+1}$, $f(x)=\frac{x^3+1}{x^2+1}$, $f(x)=\frac{2x^2-1}{3x^2+x}$; they each have different limits. (2) $f(x) = \sqrt{x^2+1}-\sqrt{x^2-1}$. (3) $f(x) = \frac{\arctan(x)-(\pi/2)}{e^{-x}}$; (4) $f(x) = 3/e^{-x}$. –  Arturo Magidin Mar 27 '12 at 19:30

3 Answers 3

up vote 1 down vote accepted

Could anyone post any examples of these indeterminations?

It's too long to evaluate in detail indeterminations of all the 4 types, with and without using L'Hôpital's rule. Just one

Example of $\infty -\infty $ indeterminations.

Sometimes these indeterminations can be evaluated without using l'Hôpital's rule. That's the case of rational fractions in $x$, i.e $\frac{P(x)}{Q(x)}$, with $P(x)$ and $Q(x)$ polynomials in $x$. As an example let $f(x)=\frac{x^{2}}{x+2}$ and $g(x)=\frac{x^{3}}{x+3}$. We have

$$\begin{eqnarray*} \lim_{x\rightarrow \infty }f(x) &=&\lim_{x\rightarrow \infty }\frac{x^{2}}{x+2}=\infty \\ \lim_{x\rightarrow \infty }g(x) &=&\lim_{x\rightarrow \infty }\frac{x^{3}}{x+3}=\infty. \end{eqnarray*}$$ Hence $\lim_{x\rightarrow \infty }f(x)-g(x)$ is indeterminate. Since we can rewrite $f(x)-g(x)$ as $$\begin{equation*} \frac{x^{2}}{x+2}-\frac{x^{3}}{x+3}=\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6}:= \frac{P(x)}{Q(x)}, \end{equation*}$$ where $P(x)=-x^{4}-x^{3}+3x^{2}$ and $Q(x)=x^{2}+5x+6$, we have $$ \begin{eqnarray*} \lim_{x\rightarrow \infty }\frac{x^{2}}{x+2}-\frac{x^{3}}{x+3} &=&\lim_{x\rightarrow \infty }\frac{P(x)}{Q(x)} \\ &=&\lim_{x\rightarrow \infty }\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6} \\ &=&\lim_{x\rightarrow \infty }\frac{-x^{2}-x+3}{1+5/x+6/x^{2}} \\ &=&\frac{\lim_{x\rightarrow \infty }-x^{2}-x+3}{\lim_{x\rightarrow \infty }1+5/x+6/x^{2}} \\ &=&\frac{-\infty }{1+0+0}=-\infty. \end{eqnarray*}$$ The polynomials $P(x)$ and $Q(x)$ are differentiable. We can thus apply l'Hôpital's rule to the fraction $$\begin{equation*} \frac{P(x)}{Q(x)}=\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6} \end{equation*}$$ as follows

$$\begin{eqnarray*} \lim_{x\rightarrow \infty }\frac{P(x)}{Q(x)} &=&\lim_{x\rightarrow \infty } \frac{P^{\prime }(x)}{Q^{\prime }(x)}\\ &=&\frac{\lim_{x\rightarrow \infty }-4x^{3}-3x^{2}+6x}{\lim_{x\rightarrow \infty }2x+5} \\ &=&\frac{\lim_{x\rightarrow \infty }-12x^{2}-6x+6}{\lim_{x\rightarrow \infty }2}=-\infty . \end{eqnarray*}$$

The final results are, of course, the same. The evaluation of

$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$

is done in this question. There are many other examples in this site.

Exercise. Try to evaluate the similar indetermination in the limit of

$$\begin{equation*} \frac{1}{x-3}+\frac{5}{\left( x+2\right) \left( 3-x\right) } \end{equation*}$$

as $x$ tends to $3$.

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The answer will generally depend on the particular $f(x)$ (though for cases 1 and 3 there is a general tool, L'Hopital's Rule if you already know it, that will often work, and for case 4 you can always say the limit does not exist).

The first limit type may be done by L'Hopital's rule (if both numerator and denominator are differentiable), or by algebraic manipulations dependent on the particular $f(x)$.

The second type of limit will require some algebraic manipulations (usually particular to the $f(x)$ in question) to bring it to some manageable form.

The third type may be done by L'Hopital's rule if both numerator and denominator are given by differentiable functions; or again by algebraic manipulations that depend on the particular form of $f$.

In the fourth case, the limit does not exist. If $n\gt 0$ and the denominator is always positive for large enough $x$, then the limit will be $\infty$; same if $n\lt 0$ and the denominator is negative for all large enough $x$. If $n\gt 0$ and the denominator is negative for all large enough $x$, or $n\lt 0$ and the denominator is always positive for large enough $x$, then the limit will be $-\infty$. Otherwise, the limit will simply not exist and not diverge to either $\infty$ or $-\infty$.

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l'Hopital's rule is a standard result that handles SOME problems of this type. It is usually taught in a Calc 1 or Calc 2 class in the United States.

If your function is in the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$, you can try to apply it right away. If your function is in the form $\infty - \infty$, you need to try to rewrite it in some way to get it in one of the other two forms.

The last form $\frac{n}{0}$ is NOT an indeterminate form.

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