Can Multivariable Limits Be Simplified Easily?

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In summary, the conversation discusses a limit with two variables and the use of L'Hopital's rule to solve it. The individual asking for help clarifies that there are only three symbols, x, a, and n, with only x being a variable. They also provide a mathematical explanation for why $x^n- a^n$ is not equal to $(x- a)(x^{n-1}- a^{n-1})$.
  • #1
goody1
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Good evening! Could anybody help me with this limit?

limit.png

I have a problem when there are two variables. The only thing I did was that:

limit2.png
,
but I don't know if it was helpful. Thank you!
 
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  • #2
What is the variable in the limit symbol?

Cbarker1
 
  • #3
goody said:
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables. The only thing I did was that:

View attachment 10129,
but I don't know if it was helpful. Thank you!

actually, factoring out $(x-a)$ in the numerator yields ...

$\dfrac{(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ... + xa^{n-2} + a^{n-1} - na^{n-1})}{(x-a)^2}$

I played with the algebra to see if I could get the other $(x-a)$ factor in the denominator to divide out ... no joy, yet.

So, starting again ...

$\displaystyle \lim_{x \to a} \dfrac{(x^n-a^n) - na^{n-1}(x-a)}{(x-a)^2}$

using L'Hopital ...

$\displaystyle \lim_{x \to a} \dfrac{nx^{n-1} - na^{n-1}}{2(x-a)}$

$\displaystyle \lim_{x \to a} \dfrac{n(x^{n-1} - a^{n-1})}{2(x-a)}$

L'Hopital again ...

$\displaystyle \lim_{x \to a} \dfrac{n(n-1)(x^{n-2})}{2} = \dfrac{n(n-1)(a^{n-2})}{2}$
 
  • #4
goody said:
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables.
No, there aren't. There are three symbols, x, a, and n. But only x is changing, it is going to a. a and n are constants. Only x is a variable.
The only thing I did was that:
View attachment 10129,
but I don't know if it was helpful. Thank you!
No. [tex]x^n- a^n[/tex] is NOT [tex](x- a)(x^{n-1}- a^{n-1}[/tex]. If you multiply the right side of that you get [tex]x^n- a^{n-1}x^n- ax^{n-1}+ a^n[/tex]. [tex]x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1}[/tex].
 
  • #5
For Country Boy, "tex" tags don't work. Dollar signs and MATH tags still work.

Quoting the last part of your post ...

No. $x^n- a^n$ is NOT $(x- a)(x^{n-1}- a^{n-1})$.
If you multiply the right side of that you get $x^n- a^{n-1}x^n- ax^{n-1}+ a^n$.

$x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1})$.
 
  • #6
Thank you. I am on entirely too many boards with too many different protocols!
 

FAQ: Can Multivariable Limits Be Simplified Easily?

What is a problem with 2 variables?

A problem with 2 variables is a mathematical or scientific problem that involves two unknown quantities or factors. These variables can be related to each other through an equation or formula, and solving the problem involves finding the values of both variables.

How do you solve a problem with 2 variables?

To solve a problem with 2 variables, you need to have at least two equations that relate the variables to each other. You can then use algebraic methods, such as substitution or elimination, to find the values of the variables that satisfy both equations. This is known as solving a system of equations.

What are some real-life examples of problems with 2 variables?

Real-life examples of problems with 2 variables can include situations where two quantities are related, such as distance and time, or cost and number of items. For instance, if you know the distance and time it takes to travel somewhere, you can use these variables to calculate the speed of travel.

What if there are more than 2 variables in a problem?

If there are more than 2 variables in a problem, it becomes more complex and may require more equations to solve. This is known as a system of equations with more than 2 variables. In these cases, it may be helpful to use methods such as Gaussian elimination or matrices to solve the problem.

Why is it important to solve problems with 2 variables?

Solving problems with 2 variables is important because it allows us to understand and make predictions about relationships between quantities. In science and mathematics, many real-world situations can be modeled using 2 variables, so being able to solve these problems is crucial for making accurate calculations and predictions.

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