2
$\begingroup$

Suppose we have to find the following limit $\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}$

Now, if we work with the De L'Hopital rule with successive differentiations we get $L=+\infty$

But if we work like this instead: $$L=\lim_{x\to\infty}\frac{x^5(1+\frac{1}{x^2}+\frac{4}{x^5})}{x^5(\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^5})}$$ then $L$ does not exist.

What is correct and what is false here? I'm a little confused.

  • 2
    In both cases you get $L=+\infty$. – 2017-02-20
  • 0
    But if we get $L=\frac{1}{0}$ doesn't this mean that the limit does not exist? – 2017-02-20
  • 1
    $\infty$ is not a number. So in both cases limit doesn't exist. – 2017-02-20
  • 0
    Thank you. So if we are asked to find this limit if it exists, then if I say $L$ doesn't exist instead of writing $L=+\infty$ I'm still correct? – 2017-02-20
  • 1
    Generally we prefer to say it has an infinite limit to distinguish from cases where it has multiple limit points. – 2017-02-20
  • 1
    @Michael You're perhaps technically correct depending on how you define the word "exists", but if I were a teacher at your level I would mark that answer wrong. You should write that the limit is positive infinity, or that the function diverges to positive infinity. – 2017-02-20
  • 0
    Many thanks. I'll keep that in mind – 2017-02-20
  • 0
    @Michael yes in you will be correct in both ways. :) – 2017-02-20
  • 0
    See my answer below, $\frac{1}{0}=undefined$ – 2017-02-20

4 Answers 4

3

First you should avoid division by 0 when calculating limit, as it will tell nothing about the limit. $\frac{1}{0}$ is not infinity, it is undefined. $$\lim_{x\to 0}{\frac{1}{x}}$$ is different scenario. Here you get limit is infinity. In your example you have to apply L'Hopital's rule $4$ times to get $$\lim_{x\to \infty}{\frac{120x}{24}}=\infty$$

1

As a side note :

Unless the sign is obvious like in $\sum\limits_{i=1}^{\infty}$ get used to explicitely write the sign of infinity $+\infty$ and $-\infty$, especially with the limit operator.

Without signing, you consider implicitely consider the two limits in $\pm\infty$. Here it happens that $\frac1x$ has the same limit $0$ in $+\infty$ and in $-\infty$ so it doesn't matter.

Yet that explains why $\frac10$ is indeterminate :

  • if $x\to0^+$ that is $x\to0$ and $x\ge0$ then $\frac1x\to+\infty$
  • if $x\to0^-$ that is $x\to0$ and $x\le0$ then $\frac1x\to-\infty$

Thus $\frac10$ has two limits and this is the definition of being indeterminate.

But if you write $\displaystyle{x=\frac1{\frac 1x}}$ then $x\to+\infty$ transforms to $\frac1{0^+}$ which is not indeterminate anymore.

Same to the other side.

1

First off, division by $0$ is undefined. Try this, $$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}=\lim_{x\to\infty}\frac{x+\frac{1}{x}+\frac{4}{x^4}}{1-\frac{1}{x}+\frac{1}{x^4}}=\infty$$ This shows more clearly the limit without resulting in $\frac{1}{0}$

-1

First of all, $\frac 1 0 = \infty$, meaning if $f(x)\to 1$ and $g(x)\to 0$ (and $g(x)$ is positive) then $\frac {f(x)} {g(x)}\to +\infty$. It's not an indeterminate form.

However, even if it were an indeterminate form, that's not a problem. Obtaining an indeterminate form doesn't mean the limit doesn't exist, it just means "the method you used doesn't work, try something else". For example, the constant function $1$ definitely has a limit as $x\to\infty$, but if I do something crazy like rewrite it as $\frac x x$, then I get the indeterminate form $\frac \infty \infty$. Every limit can be made to look like an indeterminate form if you try to calculate it in the right way.

  • 5
    1/0 has no meaning(is undefined). Infinity has meaning. 1/0 is not infinity – 2017-02-20