How Do You Solve These Limit Problems in Calculus?

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Homework Help Overview

The discussion revolves around solving limit problems in calculus, specifically focusing on evaluating limits as x approaches infinity and a specific value. The original poster presents two limit expressions that they find challenging.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the limit as x approaches infinity for the expression 10^(1/x) and discuss the behavior of 1/x in this context. They also address a limit that results in an indeterminate form (0/0) and consider factoring as a method to resolve it. Questions arise regarding the appropriateness of multiplying by x^2 versus factoring directly.

Discussion Status

Participants are actively engaging with the problems, offering various approaches and questioning the methods suggested. Some guidance has been provided regarding manipulation of the expressions, but no consensus on the final solution has been reached.

Contextual Notes

There is mention of the original poster's uncertainty with LaTeX formatting and negative powers, which may affect their understanding of the problem. The discussion also reflects differing opinions on the best method to handle the limit expressions.

Hollysmoke
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Hi, I'm having trouble with a couple limits because I'm not sure how to answer them exactly...


a)lim x-> infinity 10^1/x (sorry i don't know how to use LaTex)


b) lim x-> 3 x^-2 - 3^-2 / x-3
Now since this one is 0/0, I tried to factor it but I'm not sure HOW to factor it.
 
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Hollysmoke said:
Hi, I'm having trouble with a couple limits because I'm not sure how to answer them exactly...


a)lim x-> infinity 10^1/x (sorry i don't know how to use LaTex)
I take it you mean 10^(1/x). You could do that as 101/x in html or [tex]10^{\frac{1}{x}}[/tex] in Latex (click on that to see the code. What does 1/x go to as x-> infinity? What is 10 to that power?


b) lim x-> 3 x^-2 - 3^-2 / x-3
Now since this one is 0/0, I tried to factor it but I'm not sure HOW to factor it.
If you don't use Latex, PLEASE use parentheses. I assume you mean
(x^(-2)- 3^(-2))/(x- 3). Did you consider multiplying numerator and denominator by x^2? That gives you (1- x^2/9)/(x^2(x-3)). If you REALLY don't like fractions multiply numerator and denominator by 9= 3^2. That gives (9- x^2)/(9x^2(x-3)). Can you factor that?
 
1) As x-> infinity, 1/x = 0, therefore 10^0 = 1

2) How come you multiplied x^2? (Our teacher was teaching us that we factor if, when the limit is put in and you get 0/0, that you should factor)
 
Because it gives you something you CAN factor! You seemed to be having difficulty with the negative powers, I got rid of them.

Of course, you know that a^2- b^2= (a-b)(a+b) so you certainly could do x^(-2)- 3^(-2)= (a^(-1))^2- (3^(-1))^2= (a^(-1)- 3^(-1))(a^(-1)+ 3^(-1)) That would also work.
 
Is the final answer -2/27?
 
That looks good to me. :)
You can always check it by plugging some value of x in the expression, and if it evaluate to some value closed to your result, then it's correct.
Say, let's choose x = 2.99 (closed to 3). Plug it in and you will have: -0.07444609..., which is ver closed to -2 / 27. And you are correct.
 

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