I'm not sure which strategy I should use. The indeterminate form is $\infty - \infty$, but I have not found a way to apply l'Hospital yet. Intuitively, this limit seems to diverge because the $x\sqrt{x^2 + 1} \approx x^2$ and $x^2$ grows a lot faster than $x$. In other words, the left term dominates the right one. So I'm looking for a direction to prove the fact. It doesn't have to be l'Hospital.
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2$\begingroup$ Hint: factor out an $x$. $\endgroup$– dxivJun 18 at 1:28
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2$\begingroup$ All you need is $\sqrt{a+1} \geq \sqrt{a}$. Can you prove that? $\endgroup$– Brian MoehringJun 18 at 1:37
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$\begingroup$ I think I can prove that. The derivative of $\sqrt{x}$ is always positive, so it's an increasing function. Therefore, $\sqrt{x + 1| \geq \sqr{x}$. But I don't clearly see how that lets me to solve the problem. $\endgroup$– user1145880Jun 18 at 15:28
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$\begingroup$ For $x\ge\frac34$, we have $\sqrt{x^2+1}\ge\frac54$, therefore, $x\sqrt{x^2+1}-x\ge\frac x4$. A more interesting question would be $\lim\limits_{x\to\infty}\left(x\sqrt{x^2+1}-x^2\right)$. $\endgroup$– robjohn ♦Jun 18 at 17:51
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6 Answers
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Hint: Multiply with the quadratic conjugate i.e. multiply with $$\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$$
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$\begingroup$ Thank you! I did it. First I factored $x$ out. Then I multiplied by your quadratic conjugate. I got $x^3/(\sqrt{x^2 +1} + 1)$, which takes the form $\infty/\infty$. Then I applied l'Hospital twice getting $6x\sqrt{x^2+1}/2 + 3x^3/2\sqrt{x^2+1}$, whose term on the right is still unclear but also taking the form $\infty/\infty$, so applying l'Hospital on it alone yields $9x^2\sqrt{x^2 +1}/2$ which goes to $\infty$. So l'Hospital alone and your quadratic conjugate prove this limit diverges. Thank you so much. $\endgroup$ Jun 18 at 15:38
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1$\begingroup$ Oops. It's much easier than that. I made mistakes above for sure. Here it is again. After multiplying by the quadratic conjugate, we get $x^3/(\sqrt{x^2 +1} + 1)$. Applying l'Hospital, we end up with $3x\sqrt{x^2 + 1}$, which explodes up. $\endgroup$ Jun 18 at 16:11
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My approach would be like this:
$x\sqrt{x^2+1}-x=x(\sqrt{x^2+1}-1)\geq 2x$ when $x\geq 2\sqrt 2$. And when $x\to +\infty$, $2x\to +\infty$, hence $x\sqrt{x^2+1}-x\to +\infty$.
The question is kind of strange though because at first glance I thought it was $\sqrt{x^2+1}-x$. It would be a little more interesting: $\sqrt{x^2+1}-x=x(\sqrt{1+\frac{1}{x^2}}-1)=x(1+\frac{1}{2}\frac{1}{x^2}+o(\frac{1}{x^2})-1)=x(\frac{1}{2x^2}+o(\frac{1}{x^2}))\to 0(x\to +\infty)$.
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The expression $x\sqrt{x^2+1}-x$ can be rewritten, by factoring $x$, as $x(\sqrt{x^2+1}-1)$.
As $x \rightarrow \infty$, $\sqrt{x^2+1} \rightarrow x$ (informally, think that the $1$ will not affect on the large value inside the radical).
Therefore, the expression $x(\sqrt{x^2+1}-1)$ can be written as $x(x-1)$ [valid for $x \rightarrow \infty$].
As $x \rightarrow \infty$, $x-1 \rightarrow x$ (informally, think that subtracting $1$ will not affect on large value of $x$).
Therefore, $x(x-1)$ can be written as $x \cdot x = x^2$ [valid for $x \rightarrow \infty$].
Clearly, $x^2$ tends to infinity as $x$ tends to infinity.
$$\implies \lim_{x \rightarrow \infty} \bigg( x\sqrt{x^2+1}-x \bigg) \rightarrow \infty$$
answered Jun 18 at 4:57
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$\begingroup$ It's great that you know how use the written language to show your thought process. So much easier to read (and pretty to see). Thank you so much. $\endgroup$ Jun 18 at 15:42
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For any $x>0$, we have $$ x \sqrt{x^2+1}-x \geqslant x \sqrt{x^2}-x=x^2(x-1) \rightarrow \infty $$ Therefore $\textrm{ as } x\rightarrow \infty ,$
$$ \qquad\qquad x \sqrt{x^2+1}-x \rightarrow \infty .$$
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$\begingroup$ Very nice argument. I begin to realize what Brian Moehring had meant. $\endgroup$ Jun 18 at 15:39
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Simply,
$$\sqrt{1 + x^2} > \sqrt{x^2} = |x| = x\,, \quad x > 0 \\ \therefore x\sqrt{1 + x^2} > x^2\,, \quad x > 0 \\ x\sqrt{1 + x^2} - x > x^2 - x\,, \quad x > 0$$
$$\lim\limits_{x \to \infty}{x^2 - x} = \lim\limits_{x \to \infty}{\left( x - \dfrac{1}{2}\right)^2 - \dfrac{1}{4}} \approx \lim\limits_{x \to \infty}{x^2} \to \infty$$
$$\lim\limits_{x \to \infty}{x\sqrt{1 + x^2} - x} \to \infty$$
Alternatively, one could use logarithms, since $f(x) > \ln{f(x)}$
$$\sqrt{1 + x^2} > \sqrt{x^2} = |x| = x\,, \quad x > 0 \\\sqrt{1 + x^2} - 1 > \ln{\left(\sqrt{1 + x^2} - 1\right)} > \ln{(x - 1)}$$
Now, $\lim\limits_{x \to \infty}{\ln{(x - 1)}} \to \infty$. The rest should be clear.
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As suggested in the comments, we have that
$$ x \sqrt{x^2 + 1} - x =x\cdot \left(\sqrt{x^2 + 1} - 1\right) \to \infty \cdot \infty=\infty$$
