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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?

Sahiba Arora
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elcocodrilotito
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    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 03 '19 at 15:43
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    It might also help to know, have you started learning about derivatives/l'hopital's rule? – Kitter Catter Jun 03 '19 at 15:47
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    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. – Z Ahmed Jun 03 '19 at 16:13

5 Answers5

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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.

Matthew Leingang
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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.

J_P
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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 $f(x) \to 0$.

SlipEternal
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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 $$0$$

Dr. Sonnhard Graubner
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$$\lim_{x\to-\infty}(1+2x^2+2x\sqrt{1+x^2})$$$$=\lim_{x\to-\infty}(\sqrt{1+x^2}+x)^2$$$$=\lim_{x\to-\infty}\frac{1}{(\sqrt{1+x^2}-x)^2}$$$$=\lim_{x\to\infty}\frac{1}{(\sqrt{1+x^2}+x)^2}$$$$=0$$

Harish Chandra Rajpoot
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