Limits

[intervade]

Im having some trouble on a couple of limits.


1. Lim ln(x-1)/(x^2 - 4) as x->2 (Estimate the limit(if it exist) by using a numerical approach.)

If I plug in 2 here, you get 0/0 - This is Degenerate? What do I do with this problem now? Do I now find values close to 2 and estimate the limit?


2. lim 1/(2-2^(1/x)) as x-> 0

If I plug 0 into this equation, it doesnt work, I end up with a undefined fraction. The graph has asymptotes at x=1 - First off all, I said this function doesnt have a limit, is this correct? If so, are the asymptotes the reason for this because Im not looking at x=1, Im looking at x=0.

Any help is appreciated, thanks!

 

[galactus]

Are you familiar with L'Hopital's rule?. It can be used when one has an indeterminate form, such as 0/0

Take the derivative of the top and the derivative of the bottom. Like so:

The derivative of ln(x-1)=1/(1-x). The derivative of x^2-4 is 2x.

\(\displaystyle \frac{\frac{1}{x-1}}{2x}\)

Now, it is simply a matter of plugging in x=2 and getting 1/4 as the limit.

If you do not know L'Hopital yet, then perhaps we can go this route.

Expand, \(\displaystyle \frac{ln(x-1)}{x^{2}-4}=\frac{1}{4}\left(\underbrace{\frac{ln(x-1)}{x-2}}_{\text{limit=1}}-\overbrace{\frac{ln(x-1)}{x+2}}^{\text{limit=0}}\right)\).............[1]

The inside limit can be found by rewriting \(\displaystyle \frac{ln(x-1)}{x-2}\)

\(\displaystyle e^{\frac{ln(x-1)}{x-2}}=(x-1)^{\frac{1}{x-2}}\)

Let t=x-2 and we have:

\(\displaystyle e^{L}=\lim_{t\to 0}(t+1)^{\frac{1}{t}}\)

The limit on the right is a famous limit that equals e.

So, we have L=1

Sub that into [1], and we can see the limit is 1/4

 

[intervade]

Sorry, This class just started last week, we haven't gotten into any of that yet.

 

[galactus]

2. lim 1/(2-2^(1/x)) as x-> 0

If I plug 0 into this equation, it doesnt work, I end up with a undefined fraction. The graph has asymptotes at x=1 - First off all, I said this function doesnt have a limit, is this correct? If so, are the asymptotes the reason for this because Im not looking at x=1, Im looking at x=0.

Any help is appreciated, thanks!

\(\displaystyle \lim_{x\to 0}\frac{1}{2-2^{\frac{1}{x}}}\)

This one is undefined. Take a look at the 1/x exponent. You were correct.

 

[Deleted member 4993]

Im having some trouble on a couple of limits.

2. lim 1/(2-2^(1/x)) as x-> 0

If I plug 0 into this equation, it doesnt work, I end up with a undefined fraction. The graph has asymptotes at x=1 - First off all, I said this function doesnt have a limit, is this correct? If so, are the asymptotes the reason for this because Im not looking at x=1, Im looking at x=0.

Any help is appreciated, thanks!

The reason the function does not have a limit at x=0 is that the function is discontinuous (non-removable) at x = 0.

The value of the function - if you approach 0 from left-side(x < 0) - is +1/2

The value of the function - if you approach 0 from right-side(x>0) - is 0.

Hence the function approaches two different values at at x=0 and the limit does not exist.

 

[intervade]

Ok thanks! Any other ideas on the first one? Like I said , we havent used any of that, we havent even began derivatives.

 

[mmm4444bot]

… Estimate the limit (if it exist) by using a numerical approach …

… Do I now find values close to 2 and estimate the limit? …

Yes! Reasoning from a table of values is known as the "numerical approach".

You want to see if function values are approaching the same number, as x approaches 2 from the left and from the right.

If the function approaches a single value from both sides, then the limit exists; otherwise, it does not.

(You didn't name the function, so I'm calling it f. Also, machines are great for this type of grunt work.)

f(1.9) = 0.2701551685

f(1.99) = 0.2518881165

f(1.999) = 0.2501876304

f(1.9999) = 0.2500187512

f(1.99999) = 0.2500012500

f(1.999999) = .2500001250

f(2.01) = 0.2481379265

f(2.001) = 0.2498126301

f(2.0001) = 0.2499812513

f(2.00001) = 0.2499987500

f(2.000001) = 0.2499998750

Do you see any trends?

That's the numerical approach.

If the instructions for exercise #2 are the same, then make a table of function values as x gets closer to zero from both sides.

'

… I said … function [in exercise #2] doesnt have a limit, is this correct?

Yes, but be careful how you state this!

The function in exercise #2 has potentially infinite limits. It does not have a limit as x approaches either zero or one, however.

In other words, the given limit does not exist.