# Evaluating limits

## Homework Statement

Evaluate the following limit or explain why it does not exist:
limx→∞ 24x+1 + 52x+1 / 25x + (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
Homework Helper
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/

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 24x+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
Homework Helper
No, its how i posted it, not 24x+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?

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 thats it, just how i posted originally

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Evaluate the following limit or explain why it does not exist:
limx→∞ 24x+1 + 52x+1 / 25x + (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##.

scottdave
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
Homework Helper
Dearly Missed
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##?

Erenjaeger and scottdave
scottdave
Homework Helper
So what methods are you familiar with?

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

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

scottdave