Limit Evaluation: Explaining Existence or Non-Existence

[Erenjaeger]

Homework Statement

Evaluate the following limit or explain why it does not exist:
lim_x→∞ 2^4x+1 + 5^2x+1 / 25^x + (1/8)^6x

The Attempt at a Solution

I know there is the method where you divide through by the highest term in the denominator, but can that be applied here?

 

[scottdave]

So is it really something like this?
$$\frac {24^{x+1} + 5^{2x+1}} {25^x + (1/8)^{6x}}$$
This is my guess as what you are trying to type.
Also, is x approaching zero or infinity? It looks like a degrees symbol.
Try using LaTeX. It is not too hard to learn and it makes it easier to read.
Here is a guide. https://www.physicsforums.com/help/latexhelp/

 

[Erenjaeger]

So is it really something like this?
$$\frac {24^{x+1} + 5^{2x+1}} {25^x + (1/8)^{6x}}$$
This is my guess as what you are trying to type.
Also, is x approaching zero or infinity? It looks like a degrees symbol.
Try using LaTeX. It is not too hard to learn and it makes it easier to read.
Here is a guide. https://www.physicsforums.com/help/latexhelp/

No, its how i posted it, not 24^x+1
and x is approaching infinity just like in the original post...
But okay ill use it, I learned it for python notebooks awhile ago, just quicker to not use it

 

[scottdave]

No, its how i posted it, not 24^x+1
and x is approaching infinity just like in the original post..

Like this?
So is it really something like this?
$$\frac {2^{4x+1} + 5^{2x+1}} {25^x + (1/8)^{6x}}$$
What I was getting at is: what is in the numerator and what is in denominator?

 

[Erenjaeger]

Like this?
So is it really something like this?
$$\frac {2^{4x+1} + 5^{2x+1}} {25^x + (1/8)^{6x}}$$
What I was getting at is: what is in the numerator and what is in denominator?

yep that's it, just how i posted originally

 

[Ray Vickson]

Homework Statement

Evaluate the following limit or explain why it does not exist:
lim_x→∞ 2^4x+1 + 5^2x+1 / 25^x + (1/8)^6x

The Attempt at a Solution

I know there is the method where you divide through by the highest term in the denominator, but can that be applied here?

You wrote
$$2^{4x+1} + \frac{5^{2x+1}}{25^x} + (1/8)^{6x} $$
If you really mean
$$\frac{2^{4x+1} + 5^{2x+1}}{25^x} + (1/8)^{6x}$$
or
$$\frac{2^{4x+1} + 5^{2x+1}}{25^x + (1/8)^{6x}}$$
then you need to use parentheses. An expression like "A+B/C+D" means $A + \frac{B}{C} + D$, but "(A+B)/(C+D)" is unambiguously equal to $\frac{A+B}{C+D}$, and "(A+B)/C+D" is unambiguously equal to $\frac{A+B}{C} + D$.

 

[Erenjaeger]

You wrote
$$2^{4x+1} + \frac{5^{2x+1}}{25^x} + (1/8)^{6x} $$
If you really mean
$$\frac{2^{4x+1} + 5^{2x+1}}{25^x} + (1/8)^{6x}$$
or
$$\frac{2^{4x+1} + 5^{2x+1}}{25^x + (1/8)^{6x}}$$
then you need to use parentheses. An expression like "A+B/C+D" means $A + \frac{B}{C} + D$, but "(A+B)/(C+D)" is unambiguously equal to $\frac{A+B}{C+D}$, and "(A+B)/C+D" is unambiguously equal to $\frac{A+B}{C} + D$.

Oh you're right, I should have actually used parentheses, i'll make sure I am clearer in the future. Any help with the question though?

 

[Ray Vickson]

Oh you're right, I should have actually used parentheses, i'll make sure I am clearer in the future. Any help with the question though?

Look at the dominant terms:
We have $2^{4x+1} = 2 (2^4)^x = 2 \;16^x$ and $5^{2x+1} = 5 (5^2)^x = 5\; 25^x$. So, among the two terms in the numerator ($2\; 16^x$ and $5\; 25^x$), which dominates for $x \to \infty$? We also have $(1/8)^{6x} = (1/8^6)^x = 1/262144^x$, so which of the two terms in the denominator will dominate when $x \to \infty$?

 

[scottdave]

So what methods are you familiar with?

 

[scottdave]

Try rewriting 25 as 5^2. What is going to happen to the (1/8) term?

 

[scottdave]

Were you able to use our suggestions to arrive at an answer?

 

[scottdave]

You can use the brute force method - plug in increasingly large numbers for x to see if it is converging toward something.
I like to do this in a spreadsheet, as a check to see if I'm on the right track.

 

[Erenjaeger]

Were you able to use our suggestions to arrive at an answer?

yeah the answer is 5