$$\lim_{x\to -\infty}\ln\left(\frac{x^2+1}{x-3}\right)=\infty$$

This is the answer I get from wolfram alpha, but shouldn't the answer be the limit doesn't exist? For large negative values of x, we can ignore the +1 and -3 so we can change the limit to $$\lim_{x\to -\infty}\ln\left(\frac{x^2}{x}\right)$$ As x approaches -$\infty$, $\left(\frac{x^2}{x}\right)$ also approaches -$\infty$ so we get $\ln\left(-\infty\right)$. However, $\ln\left(-\infty\right)$ doesn't make sense because ln(x) isn't even defined for negative numbers. So, the limit doesn't exist and is therefore undefined. Am I wrong?

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    $\begingroup$ Yes, you should be right. The function isn't even defined, say, at $x=-100$. $\endgroup$ – Dzoooks Jun 4 at 23:30
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    $\begingroup$ Perhaps add the link to your Wolfram Alpha computation. $\endgroup$ – Michael Burr Jun 4 at 23:31
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    $\begingroup$ I don't know if it can be applied to limits but if we extend $\ln(z)$ to the whole complex plane we could say $\ln(-x)=\ln(-1)+\ln(x)$ so we can put the limit in the form, where $\Re(L)\to\infty$ and there are will always be an imaginary part $\endgroup$ – Henry Lee Jun 4 at 23:36
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – user532874 Jun 4 at 23:36
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    $\begingroup$ Wolfram Alpha tends to assume you are working in complex-valued functions, even if the domain of the function is suggested to be real numbers. So $\ln x$ is defined for $x$ a negative real. This also means the limit should be taken as the extended complex $\infty,$ and not the extended real $+\infty.$ $\endgroup$ – Thomas Andrews Jun 5 at 0:09

As the comments suggested that Wolfram usually assumed you are working in complex-valued functions, so that $\ln(-x) = \ln(-1) + \ln(x)$ and therefore, $\ln( -\infty ) = \infty$. So you are right that the limit doesn't make sense and shouldn't exist when we consider the function to be real-valued only.

I checked to see what they did this using the step-by-step solution option and this is what they gave me: enter image description here

Hope this help.


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