Is it "safe" to evaluate $\lim_{x \to \infty} \frac{1}{\frac{1}{x} +x}$ as follows?

Question

If I want to evaluate the following limit: $$\lim_{x \to \infty} \frac{1}{\frac{1}{x} +x},$$ is it valid to use regular arithemtic rules to come to $$\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \frac{1}{x} +\lim_{x \to \infty} x}?$$ Evaluating existing limits gives $$\frac{1}{0 +\lim_{x \to \infty} x},$$ but writing something such as $$\frac{1}{\infty}=0$$ does not seem legal, although it feels intuitive that the limit is $0$ from the first expression.

Is there a more sound way of stating this? Perhaps with different laws of limits which I seem to be lacking knowledge of? Thanks in advance for your help!

Answer 1

is it valid to use regular arithmetic rules to come to $\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \frac{1}{x} +\lim_{x \to \infty} x}\;$ ?

No, because the second term in the denominator does not have a (finite) limit, and you cannot mix "regular arithmetic rules" with infinities.

Is there a more sound way of stating this?

You could derive it (correctly) as follows:

$$ \lim_{x \to \infty} \frac{1}{\frac{1}{x} +x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{x^2} +1} = \frac{\lim_{x \to \infty} \frac{1}{x}}{\lim_{x \to \infty} \frac{1}{x^2} +1} = \frac{0}{0 + 1} = 0 $$

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[ EDIT ] Additional notes, prompted by the comments.

is it valid to use regular arithemtic rules to come to $\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \frac{1}{x} +\lim_{x \to \infty} x}?$

This would follow by the "regular arithmetic rules" from the limit properties for sums and products if all limits existed and were finite. But one of the limits is not finite in this case, so it does not follow by those rules.

If the OP is familiar with (and is allowed to use) the extended reals line, then the result would follow from the arithmetic operations for extended reals, in particular $\,\frac{1}{0 + \infty}=0\,$.

Evaluating existing limits gives $\frac{1}{0 +\lim_{x \to \infty} x}$

This is wrong in general, because it is not allowed to take the limits of only some of the expressions involved. Doing so can easily give bogus results e.g. $\,\lim \left(1+\frac{1}{n}\right)^n=\lim \left(1+0\right)^n=1\,$.

Answer 2

As explained in comments, your method is not valid because the hypotheses for the quotient and sum limit laws are not satisfied. However, the common technique of (convert to polynomials, then) factor out the big part works. \begin{align*} \lim_{x \rightarrow \infty} \frac{1}{\frac{1}{x} + x} &= \lim_{x \rightarrow \infty} \left( \frac{1}{\frac{1}{x} + x} \cdot \frac{x}{x} \right) \\ &= \lim_{x \rightarrow \infty} \frac{x}{1 + x^2} \end{align*} Here the largest power of $x$ appearing in the numerator or denominator is $2$, so factor out $x^2/x^2$. Continuing the display, \begin{align*} &=\lim_{x \rightarrow \infty} \left( \frac{x^2}{x^2} \cdot \frac{1/x}{1/x^2 + 1} \right) \\ &=\frac{\lim_{x \rightarrow \infty} 1/x}{\lim_{x \rightarrow \infty} (1/x^2 + 1)} \\ &=\frac{\lim_{x \rightarrow \infty} 1/x}{\lim_{x \rightarrow \infty} 1/x^2 + \lim_{x \rightarrow \infty} 1} \\ &= \frac{0}{0 + 1} \\ &= 0 \end{align*}

This may seem like "put something in just so we can take it and a little bit more out again", and this is a valid observation. Clearing the nested denominators is just to make it very clear where the power of $x$ is coming from. However, we can do the same thing in the original form: $$ \lim_{x \rightarrow 0} \left( \frac{x}{x} \cdot \frac{1/x}{1/x^2 + 1} \right) $$ equally arranging for the numerator and denominator to have finite limits with the denominator not having limit $0$.