# Limit Evaluation: Explaining Existence or Non-Existence

• Erenjaeger
In summary, the limit does not exist because the terms in the numerator and denominator are not equal.
Erenjaeger

## 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?

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/

scottdave said:
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

Erenjaeger said:
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?

scottdave said:
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

Erenjaeger said:

## 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
Ray Vickson said:
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?

Erenjaeger said:
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
So what methods are you familiar with?

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

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

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.

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

scottdave

## What is limit evaluation?

Limit evaluation is a mathematical concept used to determine the value of a function as its input approaches a particular value. It is often used to explain the existence or non-existence of a limit at a certain point.

## How is limit evaluation used in scientific research?

Limit evaluation is commonly used in scientific research, particularly in fields such as physics and engineering, to analyze the behavior of a system or process as it approaches certain limits. It can also be used to make predictions about the behavior of a system in extreme conditions.

## What is the difference between a limit existing and not existing?

A limit exists if the output of a function approaches a particular value as its input approaches a given point. If the output does not approach a specific value, the limit does not exist. In other words, a limit exists if the function is continuous at the given point, and does not exist if there is a discontinuity at that point.

## Can limit evaluation be used to prove the existence of a phenomenon?

Limit evaluation can be a useful tool in providing evidence for the existence of a phenomenon. By analyzing the behavior of a function as it approaches certain limits, scientists can make predictions about the behavior of a system or process and use this information to support their theories about the existence of a phenomenon.

## Are there any limitations to limit evaluation?

Limit evaluation is a powerful mathematical tool, but it does have its limitations. It can only be applied to functions that are continuous at a given point, and it cannot be used to prove the existence of a phenomenon with absolute certainty. Additionally, the accuracy of limit evaluations may be affected by factors such as measurement error and simplifying assumptions.

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