Math Test: Limits, Derivatives & Combination/Permutation

In summary, the conversation was about an upcoming math test on limits, derivatives, and combination & permutation. The person had questions about dealing with indeterminate forms and finding the limit of certain expressions. They also asked about when to study the right and left derivative. The conversation ended with a clarification on using LaTeX code.
  • #1
mohlam12
154
0
Hey!
So monday I have my math test, it's going to be about limits, derivatives and combination & permutation ...
I have some question;
if you get an indeterminate form of +[tex]\infty - \infty[/tex], is there any rule to ease everything for you. like how you can use the Hopital ruel for 0/0 ? anw what do you do in general when you get this kind of indeterminate form?
if they ask you to see if the function is derivable in [tex]\x_{0}[/tex], when do you study the right and left derivation ?
that s it for now!
hope to hear from you soon. Thanks
Mohammed
 
Physics news on Phys.org
  • #2
mohlam12 said:
Hey!
So monday I have my math test, it's going to be about limits, derivatives and combination & permutation ...
I have some question;
if you get an indeterminate form of +[tex]\infty - \infty[/tex], is there any rule to ease everything for you. like how you can use the Hopital ruel for 0/0 ? anw what do you do in general when you get this kind of indeterminate form?
Uhmm, can you be little bit clearer? Like giving us an example, or something along those lines...
If you were ask to find the limit of some radicals, then it's common to multiply and divide the whole thing by its conjugate expression.
------------
Example:
Find
[tex]\lim_{x \rightarrow +\infty} \sqrt{x ^ 2 + x} - x[/tex]
Now, it's the Indeterminate form [tex]+ \infty - \infty[/tex], right?
Multiply the whole expression with [tex]\frac{\sqrt{x ^ 2 + x} + x}{\sqrt{x ^ 2 + x} + x}[/tex] to obtain:
[tex]\lim_{x \rightarrow +\infty} \sqrt{x ^ 2 + x} - x = \lim_{x \rightarrow +\infty} \frac{(\sqrt{x ^ 2 + x} - x) (\sqrt{x ^ 2 + x} + x)}{\sqrt{x ^ 2 + x} + x}[/tex]
[tex]= \lim_{x \rightarrow +\infty} \frac{x ^ 2 + x - x ^ 2}{\sqrt{x ^ 2 + x} + x} = \lim_{x \rightarrow +\infty} \frac{x}{\sqrt{x ^ 2 + x} + x}[/tex]
Now divide both numerator and denominator by x, note that as x tends to positive infinity, then x > 0, so [tex]x = \sqrt{x ^ 2}[/tex]
[tex]\lim_{x \rightarrow +\infty} \frac{x}{\sqrt{x ^ 2 + x} + x} = \lim_{x \rightarrow +\infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{2}[/tex].
Can you get this? :)
if they ask you to see if the function is derivable in [tex]\x_{0}[/tex], when do you study the right and left derivation ?
that s it for now!
hope to hear from you soon. Thanks
Mohammed
Do you mean differentiable? Yes, kind of, if f(x0) is differentiable at x0, then we must have:
[tex]f'(x_0 ^ +) = f'(x_0 ^ -)[/tex].
------------
By the way, we don't need a slash before an x in your LaTeX code, we just need to type x_0, backslashes are used to make functions displayed in normal font (i.e not in italics) so that they'll stand out from the rest (which is in italic font).
Example: \sin x
[tex]\sin x[/tex]
 
Last edited:
  • #3


Hi Mohammed,

It sounds like you have a lot of great questions about your upcoming math test. For the indeterminate form of +\infty - \infty, there is a rule called the "limit of the difference equals the difference of the limits" which can be used to simplify the expression. However, if this rule doesn't work for your specific problem, you can also try using L'Hopital's rule, which you mentioned. This rule states that if you have a limit of the form 0/0 or \infty/\infty, you can take the derivative of the numerator and denominator separately and then evaluate the limit again. As for when to use right and left derivatives, it depends on the function and the specific point you are studying. In general, you should use the right derivative when approaching the point from the right side and the left derivative when approaching from the left side. I hope this helps and good luck on your test!
 

FAQ: Math Test: Limits, Derivatives & Combination/Permutation

What is the difference between a limit and a derivative?

A limit is the value that a function approaches as the input approaches a specific value. A derivative, on the other hand, is the instantaneous rate of change of a function at a specific point.

How do I find the limit of a function?

To find the limit of a function, you can plug in values that are approaching the desired input value and see what value the function is approaching. You can also use algebraic techniques such as factoring and simplifying to evaluate the limit algebraically.

What is the chain rule in calculus?

The chain rule is a rule in calculus that allows you to find the derivative of composite functions. It states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

How do I calculate combinations and permutations?

To calculate combinations, you use the formula nCr = n! / (r!(n-r)!), where n is the total number of objects and r is the number of objects being chosen. To calculate permutations, you use the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects being chosen in a specific order.

What are some real-world applications of limits and derivatives?

Limits and derivatives have many real-world applications, such as determining the maximum or minimum values of a function, finding instantaneous rates of change in physics and engineering problems, and optimizing business strategies in economics and finance.

Similar threads

Replies
9
Views
485
Replies
4
Views
1K
Replies
2
Views
1K
Replies
4
Views
2K
Replies
3
Views
4K
Replies
53
Views
2K
Replies
5
Views
6K
Replies
10
Views
2K
Replies
1
Views
1K
Back
Top