Using limits to solve an arithmetic problem?

[tmt1]

Supposing $x = 0$, do I need limits to solve $\frac{lnx }{x^2} + \frac{1}{2x^2}$?

Since $lnx$ does not exist at $x = 0$, then the best we can do is $\lim_{{x}\to{0}} lnx$ which is $0$.

But then, $x^2$ at zero equals 0, so we have $0/0$ which is indeterminate, so I need to find $\lim_{{x}\to{0}} \frac{ln x}{x^2}$.

So maybe I can apply l'hopital's rule here, and take the derivative of the numerator and denominator:

$\frac{\frac{1}{x}}{2x}$

and then again to get:

$\lim_{{x}\to{0}} \frac{\frac{-1}{x^2}}{2}$ which is clearly $- \infty$.

Then $\lim_{{x}\to{0}} \frac{1}{2x^2}$ is clearly positive infinity.

So then I end up with $-\infty + \infty$ which is indeterminate, so how would I resolve this?

 

[MarkFL]

I would begin by stating:

$$\frac{\ln(x)}{x^2}+\frac{1}{2x^2}=\frac{2\ln(x)+1}{2x^2}=\frac{\ln(x^2)+1}{2x^2}$$

As $x\to0$, we do not get an indeterminate form, we get $$-\frac{\infty}{0}$$, which is just $-\infty$. :D

 

[topsquark]

Supposing $x = 0$, do I need limits to solve $\frac{lnx }{x^2} + \frac{1}{2x^2}$?

It's hard to misinterpret your question but the way you have stated it is not really that good. Your expression does not exist for x = 0 plain and simple. You don't need to look at limits. If you are going to take the limit as x goes to 0 you need to mention that from the start.

-Dan

 

[tmt1]

It's hard to misinterpret your question but the way you have stated it is not really that good. Your expression does not exist for x = 0 plain and simple. You don't need to look at limits. If you are going to take the limit as x goes to 0 you need to mention that from the start.

-Dan

Interesting. I needed to solve this as part of a larger question that did not explicitly use limits. I think if I had been able to solve it without limits, it would have been satisfactory for the purposes of the problem. However, if I just stated that the expression does not exist, that would not be satisfactory.

Thanks for the input.

 

[HallsofIvy]

f(x)= \frac{x^2- 4}{x- 2} is not defined at x= 2.

I could note that, as long as x is not 2, \frac{x^2- 4}{x- 2}= \frac{(x- 2)(x+ 2)}{x- 2}= x+ 2 so that \lim_{x\to 2} f(x)= \lim_{x\to 2} x+2= 4.

So if we were to change the definition of f by defining f(2)= 4, then it would be identical to x+ 2. But that is not the way it is originally defined.