# Does this limit exist or is undefined?

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

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?

• Yes, you should be right. The function isn't even defined, say, at $x=-100$. – Dzoooks Jun 4 at 23:30
• Perhaps add the link to your Wolfram Alpha computation. – Michael Burr Jun 4 at 23:31
• 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 – Henry Lee Jun 4 at 23:36
• wolframalpha.com/input/… – user532874 Jun 4 at 23:36
• 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 $+\infty.$ – 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.