Evaluating limits

  • #1
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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?
 

Answers and Replies

  • #2
scottdave
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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/
 
  • #3
141
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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
 
  • #4
scottdave
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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?
 
  • #5
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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
 
  • #6
Ray Vickson
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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##.
 
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  • #7
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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?
 
  • #8
Ray Vickson
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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##?
 
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  • #9
scottdave
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So what methods are you familiar with?
 
  • #10
scottdave
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Try rewriting 25 as 5^2. What is going to happen to the (1/8) term?
 
  • #11
scottdave
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Were you able to use our suggestions to arrive at an answer?
 
  • #12
scottdave
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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.
 
  • #13
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Were you able to use our suggestions to arrive at an answer?
yeah the answer is 5
 
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