17
$\begingroup$

When can I move $\lim$ inside an expression? what are the requirements from the function?

For example: $\displaystyle \lim_{x \to \infty}\frac{\sqrt{x^2+2}}{3x-6}$

$\endgroup$

4 Answers 4

19
$\begingroup$

By definition of the continuous function:

Function is continuous at $x_0$ iff $\lim_{x\to x_0} f(x)$ exists and

$$\lim_{x\to x_0} f(x) = f(x_0)$$


Thus, if function $f$ is continuous at $g$ and $\lim\limits_{x\to x_0} g(x)=g$ then:

$$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0} g(x)\right)$$

When you are operating on $\pm\infty$ you can flip the function inside out by substitution $x=\tfrac1u$ so that you are analyzing continuity at $u=0$.

$\endgroup$
4
$\begingroup$

The function has to be continuous. Since continuity at infinity is a controversial concept, change it to a more comfortable situation by setting $x=1/y$.

$\endgroup$
2
  • $\begingroup$ So first I need to check if the function is continuous? That mean that the limit at the point is finite? $\endgroup$
    – gbox
    Nov 8, 2015 at 11:07
  • 1
    $\begingroup$ Preferably you find a transformation so that the limit is at a finite point. Then simplify the expression so that you can apply the arithmetic laws of the limit, and the additional one that $\lim_{x\to x_0}f(x)=f(\lim_{x\to x_0}x)=f(x_0)$ whenever $f$ is continuous in $x_0$. (One-sided limits by restriction on domain.) $\endgroup$ Nov 8, 2015 at 11:12
1
$\begingroup$

You can do this, if both the limits exist. In this case both of them are $\infty$, so they don't exist and you can't move the $\lim$ inside.

In case you need help to solve this limit, try dividing numerator and denominator by $x$ or use l'Hopital's rule if you are familiar with it.

$\endgroup$
1
  • $\begingroup$ What you've said here has an important difference from other answers. Others have said the inner one has to be continuous at g and g continuous at x0. But you only say the must have limits. Is there a theorem for this you can give me the link to, or somewhere in a math book? because it's impossible to say which is correct without reference. $\endgroup$
    – aderchox
    Dec 8, 2020 at 18:43
1
$\begingroup$

Your example:

$$\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{3x-6}=$$ $$\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{3x}=$$ $$\lim_{x\to\infty}\frac{\sqrt{x^2}}{3x}=$$ $$\frac{1}{3}\lim_{x\to\infty}\frac{\sqrt{x^2}}{x}=$$ $$\frac{1}{3}\lim_{x\to\infty}\frac{x}{x}=$$ $$\frac{1}{3}\lim_{x\to\infty}1=\frac{1}{3}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.