- #1
DocZaius
- 365
- 11
Hello,
I am learning Calculus I and doing fine so far. I arrived at a spot that I can intuitively understand, but would like a more formal mathematical understanding of.
It concerns limits of the arguments of functions.
Consider:
[tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})[/tex]
Now, intuitively I can tell that the argument will go to 0 from the right, and that thus the natural log will never have a negative argument. The argument will approach 0, meanings the natural log will approach [tex]-\infty[/tex]
All that is fine, but my problem is that after I show that:
[tex]\displaystyle\lim_{x\to-\infty}\frac{1}{x^2-4x}=0[/tex]
I don't know how to mathematically (rather than reasoning through it) show the next step, the one before this final one:
[tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})=-\infty[/tex]
Any help? I might guess that maybe the step would look like this:
[tex]\displaystyle\lim_{\alpha\to0+}ln(\alpha)=-\infty[/tex]
In this (almost surely wrong) attempt at creating an intermediate step, I take the fact that the argument approaches 0 from the right and call the argument [tex]\alpha[/tex]. This transforms the original problem, but yields the correct answer. Anyways, this was just to show my thoughts. What is the correct intermediate step?
Thanks!
I am learning Calculus I and doing fine so far. I arrived at a spot that I can intuitively understand, but would like a more formal mathematical understanding of.
It concerns limits of the arguments of functions.
Consider:
[tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})[/tex]
Now, intuitively I can tell that the argument will go to 0 from the right, and that thus the natural log will never have a negative argument. The argument will approach 0, meanings the natural log will approach [tex]-\infty[/tex]
All that is fine, but my problem is that after I show that:
[tex]\displaystyle\lim_{x\to-\infty}\frac{1}{x^2-4x}=0[/tex]
I don't know how to mathematically (rather than reasoning through it) show the next step, the one before this final one:
[tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})=-\infty[/tex]
Any help? I might guess that maybe the step would look like this:
[tex]\displaystyle\lim_{\alpha\to0+}ln(\alpha)=-\infty[/tex]
In this (almost surely wrong) attempt at creating an intermediate step, I take the fact that the argument approaches 0 from the right and call the argument [tex]\alpha[/tex]. This transforms the original problem, but yields the correct answer. Anyways, this was just to show my thoughts. What is the correct intermediate step?
Thanks!