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I am trying to evaluate $$ \lim\limits_{x \to \infty} (\sqrt{x^2 + 4x} - x) $$ Multiplying by the conjugate gives a limit of $2$, but I have to apply L'Hospital's rule.

I can't even figure out how to get started and writing this in an indeterminate form, because it's currently of the form $\infty - \infty$.

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3 Answers 3

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HINT: Factoring $x^2$ out of the radicand gives

$$\lim_{x \to \infty} \sqrt{x^2\left(1+\frac{4}{x}\right)}-x=\lim_{x \to \infty} x\left(\sqrt{1+\frac{4}{x}}-1 \right)$$

You now have $\infty \times 0$. How can you make this into the proper form for L'Hopital's Rule?

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Conjucation still can be usable. A little trick at the end of the solution, I think, gives this variant the right to participate in the answers parade $$\lim\limits_{x \to \infty} (\sqrt{x^2 + 4x} - x) = \lim\limits_{x \to \infty} \frac{(\sqrt{x^2 + 4x} - x)(\sqrt{x^2 + 4x} + x) }{\sqrt{x^2 + 4x} + x}=\\ =\lim\limits_{x \to \infty} \frac{4x }{\sqrt{x^2 + 4x} + x}= \text{L'Hôpital} =\lim\limits_{x \to \infty} \frac{4 }{\frac{x+2}{\sqrt{x^2 + 4x}} + 1}$$ For fraction in denominator, assuming that limit exists $$L=\lim\limits_{x \to \infty}\frac{x+2}{\sqrt{x^2 + 4x}}= \text{L'Hôpital} =\lim\limits_{x \to \infty}\frac{1}{\frac{x+2}{\sqrt{x^2 + 4x}}}=\frac{1}{L}$$ Now, getting $L=1$ finishes example.

Addition.

Existence of $L=\lim\limits_{x \to \infty}\frac{x+2}{\sqrt{x^2 + 4x}}$ is easily seen from its bounded $\frac{x+2}{\sqrt{x^2 + 4x}}\leqslant \frac{x+2}{\sqrt{x^2}}\lt 2$ in some neighbourhood of infinity and monotone from sign of derivative.

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  • $\begingroup$ @lone student. In the OP it is written that the use of the conjugate gives the answer immediately, without the use of lopital, but there is no prohibition on the conjugate. I showed how to apply the exactly L'Hôpital to the conjugate one. $\endgroup$
    – zkutch
    Jul 19, 2021 at 6:28
  • $\begingroup$ Yes. I thought otherwise. $\endgroup$
    – lone student
    Jul 19, 2021 at 6:41
  • $\begingroup$ Who tells you that your $L$ exists? You’re only proving that if it exists, then it equals $1$ $\endgroup$
    – egreg
    Jul 19, 2021 at 7:58
  • $\begingroup$ @egreg Yes. I wanted to leave opportunity to set this question for OP. Ok. "who tells you that" it don't exist? On one hand same can be asked for initial expression. Right? But, anyway, to satisfy rigorous approach we can easily use bounded and monotony. $\endgroup$
    – zkutch
    Jul 19, 2021 at 8:08
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The best way is using the conjugates. But, when using L'Hôpital's rule we must be careful not to complicate our work further.

Without using conjugates, maybe we can consider the following way:

Let $x=\frac 1t$ as $t\to 0^{+}$ (we interpret $x\to\infty$ as $x\to +\infty$) you get

$$\begin{align}\sqrt{x^2+4x}-x=\sqrt{\frac {1}{t^2}+\frac 4t}-\frac 1t\\ \end{align}$$

Then, we have

$$\begin{align}\lim\limits_{x \to \infty} \left(\sqrt{x^2 + 4x} - x\right)&=\lim_{t\to 0^{+}}\sqrt{\frac {1}{t^2}+\frac 4t}-\frac 1t\\ &=\lim_{t\to 0^{+}}\sqrt{\frac {4t+1}{t^2}}-\frac 1t\\ &=\lim_{t\to {0^+}}\frac{\sqrt{4t+1}-1}{t}\\ &=\lim_{t\to 0^{+}}\frac{2}{\sqrt{4t+1}}\\ &=2.\end{align}$$

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  • $\begingroup$ Technically, the limit should be $t \rightarrow 0^+$, but then when you find the more general limit $t \rightarrow 0$ that gives you that one for free. $\endgroup$
    – David
    Jul 19, 2021 at 4:58
  • $\begingroup$ Would you mind explaining why you were able to take the positive root of $t$? This is the only step I don't fully understand. $\endgroup$
    – Mathematical Rookie
    Jul 19, 2021 at 4:58
  • $\begingroup$ @Algebra $x\to +\infty$ means $t>0$. $\endgroup$
    – lone student
    Jul 19, 2021 at 5:38
  • $\begingroup$ @David I accepted $t>0,$ because $x\to +\infty$ . Now, moved to the "official notation". $\endgroup$
    – lone student
    Jul 19, 2021 at 6:48
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    $\begingroup$ This is somewhat circular, because you're essentially computing the derivative of $f(t)=\sqrt{4t+1}$ at $0$ using the derivative of $f$. $\endgroup$
    – egreg
    Jul 19, 2021 at 8:22

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