# Why do we solve lim in two different ways?

When solving the limit in this form $$\lim_{x \to \infty} \sqrt{x} \pm \sqrt{y}$$ where $$x$$ and $$y$$ are two different expressions, I was taught that I should first find the element with the highest degree in each of $$x$$ and $$y$$, then if they share the same degree and same coefficient I should solve it by multiplying and dividing by the conjugate of the provided expression. else I should I factor as usual.

here is an example limit $$\lim_{x \to \infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x+1}\\$$ This is my attempt $$\lim_{x \to \infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x+1}\\ \lim_{x \to \infty} \sqrt{x^2} - \sqrt{x^2}\\ \lim_{x \to \infty} x-x\\ =0$$ (My thinking with removing the $$x+1$$ and $$-3x+1$$ was that the $$x^2$$ overruled them if that's the right word )

where as my professor did the following: $$\lim_{x \to \infty} \dfrac{\sqrt{x^2+x+1} - \sqrt{x^2-3x+1} \times \sqrt{x^2+x+1} + \sqrt{x^2-3x+1}} {\sqrt{x^2+x+1} + \sqrt{x^2-3x+1}}\\ \lim_{x \to \infty} \dfrac{x^2+x+1 - x^2-3x+1} {\sqrt{x^2+x+1} + \sqrt{x^2-3x+1}}\\ \lim_{x \to \infty} \dfrac{4x} {2x}\\ = 2$$

when I questioned about it, my professor explained that I should always use the conjugate when the $$x$$ has the same degree and coefficient in each root (as is the case in our problem)

in conclusion my question is why the two different methods, depending on the different $$x$$ coefficient and why is my attempt wrong

• In your first attempt, consider doing the same thing when the expressions don't have a square root (i.e. change the exponents from $\frac{1}{2}$ to $1).$ The 1st method is a short-hand method for a procedure in which you can rewrite all terms in a QUOTIENT by dividing by the highest power of $x$ that appears anywhere in the quotient (this doesn't change what you're taking the limit of), then use rules like "limit of quotient equals quotient of limits" (when each of the new numerator and denominator limits exist). The 2nd method above also doesn't change what you're taking the limit of. Commented Dec 5, 2023 at 21:01
• Your initial $\lim_{x \to \infty} \sqrt{x} \pm \sqrt{y}$ does not suggest you are considering two different functions of $x$ Commented Dec 5, 2023 at 21:07
• You have implicitly assumed $\lim\limits_{x \to \infty} \big(\sqrt{x^2+x+1} - \sqrt{x^2-3x+1}\big) = \lim\limits_{x \to \infty} \big(\sqrt{x^2+x+1}\big) - \lim\limits_{x \to \infty} \big(\sqrt{x^2-3x+1}\big)$ but that would give you the undefined $\infty-\infty$ Commented Dec 5, 2023 at 21:10
• @Henry okay, I did more research and it seems that my attempt is wrong because it leads to an undetermined form as you said. I guess than my question becomes, why does using a conjugate get rid of this undetermined form? Commented Dec 5, 2023 at 21:12
• You are taking advantage of $a-b=\frac{a+b}{a^2-b^2}$ where in this case $a^2-b^2$ is simpler than $a-b$ and where $a+b$ involves addition rather than subtraction of two terms of similar size Commented Dec 5, 2023 at 21:23

I'm currently studying calculus 1 and for what I've been taught you cannot just ignore, for example, $$x+1$$ in the first square root just because $$x^2$$ goes to infinity "faster" than $$x+1$$; when you find an indeterminate form $$[\infty-\infty]$$ which comes from two squared roots you should always try to rationalize by multiplying numerator and denominator for the conjugate to said irrational term (it clearly is equivalent to multiplying by one and therefore you don't change the outcome of the limit, if it exists, while solving the indeterminate form issue).

• Yes, you're essentially right here. You can't ignore the $x+1$ because, after the cancellation $x^2-x^2=0$, that's the critical part that's left (along with the corresponding part from the other radical as well) to determine the limit! The idea is $$(\textrm{Big thing}+A)-(\textrm{Big thing}+B) = A - B$$ so of course you can't ignore $A$ and $B$. +1 for you.
– MPW
Commented Dec 5, 2023 at 21:19

You're probably borrowing this technique from a similar context, where it is valid to throw away lower-order terms: when you're computing the limit of a ratio of two terms $$\lim_{x \to \infty} \frac {\sqrt{x^2 + x + 1}}{\sqrt{x^2 - 3x + 1}} = \lim_{x \to \infty} \frac {\sqrt {x^2}}{\sqrt {x^2}} = 1$$

What you have here is not a ratio of expressions, but a difference. In the former case, the total change in the expression, from say $$\frac {\sqrt{x^2 + x + 1}}{\sqrt{x^2 - 3x + 1}}$$ to $$\frac {\sqrt{(x+1)^2 + (x+1) + 1}}{\sqrt{(x+1)^2 - 3(x+1) + 1}}$$, depends on the relative difference between $$x$$ and $$y$$ (since the square root distributes over the fraction), so the higher-order term $$x^2$$ dominates.

In the case that you have a difference, $$\sqrt x \pm \sqrt y$$ rather than $$\dfrac{\sqrt x}{\sqrt y}$$. The difference between these two terms will be proportionally controlled by the $$x^2$$ terms, but because they're not being divided by a large number as in the ratio case, the error introduced by discarding lower order terms will not actually approach zero, so you can't discard it in the limit.

To solve this problem, your teacher used conjugation to write this expression as a ratio, at which point you are able to discard higher order terms (as they did) and arrive at the correct answer.

As a minor other note, you're missing parentheses in your teacher's solution. The first line should read

$$\lim_{x \to \infty} \frac{\left(\sqrt{x^2 + x + 1} + \sqrt{x^2 - 3x + 1}\right) \times \left(\sqrt{x^2 + x + 1} - \sqrt{x^2 - 3x + 1}\right)}{\sqrt{x^2 + x + 1} + \sqrt{x^2 - 3x + 1}}$$