Both these limits tend to infinity but it is obvious to say that
$$\lim_{x \to 0}\frac{2}{x} \gt \lim_{x \to 0}\frac{1}{x}$$
as at any point it is true, if not what about these one
$$\lim_{x \to 0}\frac{1}{x^2} \gt \lim_{x \to 0}\frac{1}{x}$$
$$\lim_{x \to 0}\frac{1}{x^3} \gt \lim_{x \to 0}\frac{1}{x}$$
If all infinities are equal then all graphs mus intersect at a point which is never true
Can one infinity be greater than other?
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Depending on your concept of infinity, yes, one infinity can be greater than another. However, when taking limits of real-valued functions, there really is only one infinity. That infinity means "The function grows beyond any finite bound", and that's it. There is, in this respect, no notion of how fast it grows beyond any finite bound; they all reach the same $\infty$ in the end.
One case where infinities have different sizes are when they measure the size of sets. The infinity which describes the number of elements in the set of natural numbers is strictly smaller than the infinity which describes the number of elements in the set of the real numbers.
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Infinities can be compared by calculating $$\lim_{x \to a} \frac{f\left(x\right)}{x}, f\left(a\right) = \infty$$
if they are also $\infty$ then calculate
$$\lim_{x \to a} \frac{f\left(x\right)}{x^2}, f\left(a\right) = \infty$$
and so on this can compare two infinities
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No.
Infinity is a concept, not a number. Therefore, the limits, while one accelerates at a faster rate, both have the same endpoint of infinity, and therefore are equal.
Have a look at this for a clearer explanation of why $\infty$ is not a number.
As already noted $\infty$ is used in limits as a symbolic manner to express that the function becomes larger than any fixed bound $M$ as $x\to 0$ (or smaller for the case $-\infty$).
What we can to compare two different functions is to consider their ratio for a same value of $x$ and also take the limit of that ratio, that is for example
then
$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{\frac1{x^2}}{\frac1{x^4}}=\lim_{x \to 0} \frac{x^4}{x^2}=\lim_{x \to 0} x^2=0$$
then we say that $g(x)$ tends to $\infty$ faster than $f(x)$ for $x \to 0$.