# When can I move the limit operand into a function?

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

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

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 $$u=0$$.

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

• So first I need to check if the function is continuous? That mean that the limit at the point is finite?
– gbox
Nov 8, 2015 at 11:07
• 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.) Nov 8, 2015 at 11:12

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.

• 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. Dec 8, 2020 at 18:43

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