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

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

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By definition of the continuous function:

Function is continuous at $x_0$ if $\lim_{x\to x_0} f(x)$ 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 $u=0$.

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

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  • $\begingroup$ 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 '15 at 11:07
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    $\begingroup$ 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.) $\endgroup$ – LutzL Nov 8 '15 at 11:12
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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.

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Your example:

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

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