1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 $1+2x^2$? $\endgroup$ – Kitter Catter Jun 3 '19 at 15:43
  • 1
    $\begingroup$ It might also help to know, have you started learning about derivatives/l'hopital's rule? $\endgroup$ – Kitter Catter Jun 3 '19 at 15:47
  • 1
    $\begingroup$ 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. $\endgroup$ – Dr Zafar Ahmed DSc Jun 3 '19 at 16:13
2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $f(x) \to 0$.

$\endgroup$
1
$\begingroup$

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 $$0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.