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

I got this limit which gives me $\ln(0^+)=-\infty$. Is this ok? I ended up with my answer in this way: my limit is equal to $$\ln\left(\lim_{x\to-\infty}\left(\frac{4}{\sqrt{x^2+4}-x}\right)\right)=\ln0^+=-\infty$$

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Let $-1/x=h$

$F=\sqrt{\dfrac{1+4h^2}{h^2}}-\dfrac1h=\dfrac{\sqrt{1+4h^2}-1}h=\dfrac{4h^2}{h(\sqrt{1+4h^2}+1)}$

$\lim_{h\to0^+}F=0$

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Perhaps you can show us some of your work; but it looks like you are claiming that $\infty - \infty = 0$, which is wrong. Such is an indeterminate form and further work would be required.

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Hints: Your answer is correct. To rule out that you have made any mistakes in the calculation: one way to calculate this limit is 1. Using that the logarithm is continuous, so you can "pass the limit through the logarithm" 2. For evaluating the expression inside the logarithm multiply numerator and denominator by$\sqrt{x^2+4} - x$.

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I ended up with my answer in this way: my limit is equal to $$\ln(\lim_\limits{x\to-\infty}(\frac{4}{\sqrt{x^2+4}-x}))=\ln0=-\infty$$

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    $\begingroup$ That is correct. Please don't post it as an answer but rather put it into your question body. $\endgroup$ – Viktor Glombik Jun 24 at 8:00
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    $\begingroup$ And whatever you do: please don't even write $\;\log 0\;$ ...it looks awful ! $\endgroup$ – DonAntonio Jun 24 at 8:14

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