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I have to find this limit $\lim_{x\to\infty}\ln(e^x+6x)\left(x-\sqrt{x^2+4}\right)$. I tried to simplify by using the properties of the logarithm and rationalizing the square root. What I get is:

$$\lim_{x\to\infty}\left(\ln(e^x)+\ln \left(1+\frac{e^x}{6x}\right)\right)\left(\frac{-4}{x+\sqrt{x^2+4}}\right).$$

But I don't know how to further simplify this problem. Could someone help me? Thanks.

M. A.
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pepper
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    $\log(e^x+6x)=\log(e^x(1+6x e^{-x}))=x+\log(1+6xe^{-x})$ – Sine of the Time Jan 24 '25 at 16:57
  • Your $e^x/(6x)$ in your first statement is backwards. It should be $6x/e^{x}.$ – Thomas Andrews Jan 24 '25 at 16:59
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    Some intuition for large $x$: $\ln(e^x + 6x) \approx \ln(e^x) = x$, this explains how to arrive at the help provided in the first comment. Do something similar for $x - \sqrt{x^2+4}$ to get a good hint of the final answer. – Villa Jan 24 '25 at 17:18

4 Answers4

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Focusing on the sub-expression $y := x-\sqrt{x^2+4}$,

$$y(x+\sqrt{x^2+4}) = (x-\sqrt{x^2+4})(x+\sqrt{x^2+4})$$ $$y(x+\sqrt{x^2+4}) = -4$$

For sufficiently large values of $x$, $\sqrt{x^2+4} \approx x$, so:

$$y(2x) \approx -4$$ $$y \approx -\frac{2}{x}$$

(WolframAlpha plot of $y$ versus its large-$x$ approximation.)

And per @Villa's comment, $\ln(e^x+6x) \approx \ln(e^x) = x$.

Combining these two factors:

$$\ln(e^x+6x)(x-\sqrt{x^2+4}) \approx \left(x\right)\left(-\frac{2}{x}\right) = -2$$

Edit: With a bit more formality,

$$L := \lim_{x\to\infty}\ln(e^x+6x)(x-\sqrt{x^2+4})$$ $$= \lim_{x\to\infty} \frac{\ln(e^x+6x)(x-\sqrt{x^2+4})(x+\sqrt{x^2+4})}{x+\sqrt{x^2+4}}$$ $$= \lim_{x\to\infty} \frac{\ln(e^x+6x)(-4)}{x+\sqrt{x^2+4}}$$

This has the form $\frac{-\infty}{+\infty}$, so apply L'Hôpital's Rule.

$$L = -4\lim_{x\to\infty} \frac{\frac{1}{e^x+6x}(e^x + 6)}{1+\frac{1}{2}(x^2+4)^{-1/2}(2x)}$$ $$L = -4\lim_{x\to\infty} \frac{\frac{e^x + 6}{e^x+6x}}{1+\frac{x}{\sqrt{x^2+4}}}$$ $$L = -4 \cdot \frac{1}{1+1} = -2$$

Dan
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It is better to write the expression as $(x - \sqrt{x^2 + 4})\ln(e^x + 6x)$ to eliminate the ambiguity.

This type of problems is best dealt with Taylor's expansion with Peano's remainder, as below: \begin{align*} x - \sqrt{x^2 + 4} &= x - x\left(1 + \frac{4}{x^2}\right)^{1/2} \\ &= x - x\left(1 + \frac{1}{2}\frac{4}{x^2} + o(x^{-2})\right) \\ &= -\frac{2}{x} + o(x^{-1}). \\ \ln(e^x + 6x) &= \ln(e^x(1 + 6xe^{-x})) = x + \ln(1 + 6xe^{-x}) \\ &= x + 6xe^{-x} + o(xe^{-x}). \end{align*} Multiplying them together yields \begin{align*} & (x - \sqrt{x^2 + 4})\ln(e^x + 6x) \\ =& -2 - 12e^{-x} + o(e^{-x}) + o(1) + o(e^{-x}) + o(e^{-x}) \\ =& -2 + o(1) \end{align*} as $x \to \infty$. This means \begin{align*} \lim_{x \to \infty}(x - \sqrt{x^2 + 4})\ln(e^x + 6x) = -2. \end{align*}

Zhanxiong
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Using the fact that the limit of the sum is the sum of the limits (when both exist and are finite) you have: \begin{align} \lim_{x\to \infty} (x-\sqrt{x^2+4})\log(e^x+6x)&=-\lim_{x\to +\infty}\frac{4}{x+\sqrt{x^2+4}}(\log(e^x)+\log(1+6xe^{-x}))\\&=-\lim_{x\to +\infty}\frac{4x}{x+\sqrt{x^2+4}}-\lim_{x\to +\infty}\frac{\log(1+6xe^{-x})}{x+\sqrt{x^2+4}}\\&=-4\lim_{x\to \infty}\frac{x}{x+\sqrt{x^2+4}}+0 \\&=-4\lim_{x\to \infty}\frac{1}{1+\sqrt{1+4/x^2}}=-2 \end{align}

Sine of the Time
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Simple estimates suffice. We have $$\sqrt{x^2+4}-x={4\over \sqrt{x^2+4}+x}$$ As $x\le \sqrt{x^2+4}\le x+2$ we get $${2\over x+1}\le \sqrt{x^2+4}-x\le {2\over x}$$ Moreover $$x\le \ln(e^x+6x)\le \ln (7e^x)=x+\ln 7$$ Therefore $${2x\over x+1}\le (\sqrt{x^2+4}-x)\ln(e^x+6x)\le {2(x+\ln 7)\over x}$$ Thus the limit of $(\sqrt{x^2+4}-x)\ln(e^x+6x)$ is equal $ 2.$ Hence the limit in OP is equal $-2.$

Ryszard Szwarc
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