# Problematic limit

Consider the function $$f(x)=1+2x^2+2x\sqrt{1+x^2}$$

I want to find the limit $$f(x\rightarrow-\infty)$$

We can start by saying that $$\sqrt{1+x^2}$$ tends to $$|x|$$ when $$x\rightarrow-\infty$$, and so we have that $$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\infty}(1+2x^2-2x^2)=1$$

However, if you plot the function in Desmos or you do it with a calculator, you will find that $$f(x\rightarrow-\infty)=0$$

What am I missing?

• When you take the limit on the square root you are throwing away terms that might be O(1). like you are saying $\sqrt{1+x^2} \approx |x|$, but why not do the same for $1+2x^2$? – Kitter Catter Jun 3 '19 at 15:43
• It might also help to know, have you started learning about derivatives/l'hopital's rule? – Kitter Catter Jun 3 '19 at 15:47
• Always remember that $\sqrt{x^2}= |x|$, so J_P did it right $\sqrt{1+x^2}=-x \sqrt{1+1/x^2},.$ when $x \sim -\infty$.. This is the most crucial part ot this question. Also see that his solution reasonably short. – Dr Zafar Ahmed DSc Jun 3 '19 at 16:13

For negative $$x$$, we have $$\sqrt{1+x^2}=-x\sqrt{1+\frac{1}{x^2}}=-x\left(1+\frac{1}{2x^2}+O(x^{-4})\right)$$ So we have $$\lim_{x\rightarrow-\infty}(1+2x^2+2x\sqrt{1+x^2})=\lim_{x\rightarrow-\infty}\left(1+2x^2-2x^2\left(1+\frac{1}{2x^2}+O(x^{-4})\right)\right)=\\ =\lim_{x\rightarrow-\infty}O(x^{-2})=0$$ You were missing a constant term in your approximation for $$\sqrt{1+x^2}$$, it can sometimes be dangerous to just reason "by feeling" like you seem to have done, I suggest using Taylor series for rigorous derivations in such cases.

We can start by saying that $$\sqrt{1+x^2}$$ tends to $$|x|$$ when $$x\rightarrow-\infty$$,

Not exactly. Both $$\sqrt{1+x^2}$$ and $$|x|$$ tend to $$\infty$$ as $$x \to -\infty$$. I think what you mean that each is asymptotic to the other, because their ratio tends to $$1$$ as $$x\to-\infty$$. But that doesn't mean you can replace one with the other, as we can see here.

As $$x\to -\infty$$, this limit takes the form $$\infty - \infty$$, which is an indeterminate form. One way to evaluate such a limit is to rewrite it in a form which is not indeterminate. For convenience, let's first write it as $$\lim_{x\to\infty} \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right)$$ Here we are just substituting $$-x$$ for $$x$$; it doesn't change the sign of $$x^2$$ or $$\sqrt{1+x^2}$$, but it does change the sign of $$x$$. The typical algebraic trick for dealing with radicals such as this is to multiply and divide by the conjugate radical: \begin{align*} 1 + 2x^2 - 2 x\sqrt{1+x^2} &= \left(1 + 2x^2 - 2 x\sqrt{1+x^2}\right) \frac{1 + 2x^2 + 2 x\sqrt{1+x^2}}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\ &= \frac{\left((1 + 2x^2) - 2 x\sqrt{1+x^2}\right)\left((1 + 2x^2) + 2 x\sqrt{1+x^2}\right)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\ &= \frac{(1+2x^2)^2 - \left(2x \sqrt{1+x^2}\right)^2}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\ &= \frac{(1+4x^2+4x^4) - 4x^2(1+x^2)}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\ &= \frac{1}{1 + 2x^2 + 2 x\sqrt{1+x^2}} \\ \end{align*} Now as $$x\to\infty$$, the denominator tends to $$\infty$$ while the numerator is a constant $$1$$. Therefore the quotient, and the original expression, tends to zero.

Multiply top and bottom by a conjugate:

\begin{align*}f(x) & = \dfrac{1+2x^2+2x\sqrt{1+x^2}}{1}\cdot \dfrac{1+2x^2-2x\sqrt{1+x^2}}{1+2x^2-2x\sqrt{1+x^2}} \\ & = \dfrac{1}{1+2x^2-2x\sqrt{1+x^2}}\end{align*}

Now, as $$x \to -\infty$$, $$1+2x^2-2x\sqrt{1+x^2} \to \infty$$, so $$f(x) \to 0$$.

Hint: We have $$f(x)=x^2\left(\frac{1}{x^2}+2-2\sqrt{1+\frac{1}{x^2}}\right)$$ So our searched limit is $$0$$