Calculate the following limits Part 1

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However, in that case you have to take the derivative of both the nominator and denominator, not just one of them.
  • #1
bugatti79
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Homework Statement



[itex]lim \underset{x \rightarrow }{\infty} {\frac{1+t^2}{1-t^2}}i+tan^{-1}(t)j+\frac{1-e^{-2t}}{t}k[/itex]

First term = [itex]=\frac{1/t^2+1}{1/t^2-1}=-1[/itex]

How do I show the second term and also the e^(-2t) term?

Thanks
 
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  • #2
Hi bugatti79! :smile:

What is the range of [itex]\tan^{-1}[/itex]?
If you don't know, you could for instance look it up in the definition of the tangent:
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

Same for [itex]e^{-2t}[/itex].
What is the smallest value that it can take (but just can't reach)?
 
  • #3
bugatti79 said:

Homework Statement



[itex]lim \underset{x \rightarrow }{\infty} {\frac{1+t^2}{1-t^2}}i+tan^{-1}(t)j+\frac{1-e^{-2t}}{t}k[/itex]
I don't understand the question. Is the limit as t --> infinity? You have x.
bugatti79 said:
First term = [itex]=\frac{1/t^2+1}{1/t^2-1}=-1[/itex]
If the limit is as t gets large, then yes.
bugatti79 said:
How do I show the second term and also the e^(-2t) term?
What do tan-1(t) and e-2t approach as t gets large?
 
  • #4
bugatti79 said:
[itex]lim \underset{x \rightarrow }{\infty} {\frac{1+t^2}{1-t^2}}i+tan^{-1}(t)j+\frac{1-e^{-2t}}{t}k[/itex]

For this limit, use [ itex] \lim_{t \to \infty} ... [ /itex ]
 
  • #5
Mark44 said:
For this limit, use [ itex] \lim_{t \to \infty} ... [ /itex ]

Ok. How about this one.

I know the sin (t) as t tends to 0 = 0 but then how is the (sin (t))/t as t tends to 0 =1?

Does y tend to 0 faster than sin(t) tends to 0?

Similarly for cos(t) / t...? Is the converse true, ie cos (t) tends to 0 faster than y tends to 0.

I.L.S, that is a good table in that wiki link. THanks.
 
  • #6
bugatti79 said:
Ok. How about this one.

I know the sin (t) as t tends to 0 = 0 but then how is the (sin (t))/t as t tends to 0 =1?

Does y tend to 0 faster than sin(t) tends to 0?

Similarly for cos(t) / t...? Is the converse true, ie cos (t) tends to 0 faster than y tends to 0.

I.L.S, that is a good table in that wiki link. THanks.

Thanks!

You're moving on to a more complex limit.
Does that mean you got the other limits?

To find
[tex]\lim\limits_{t \to 0} {\sin t \over t}[/tex]
you should apply l'Hôpital's rule:
http://en.wikipedia.org/wiki/L'Hôpital's_rule
The article gives your limit as an example.

To answer your question, sin t and t approach 0 with equal speed.
 
  • #7
bugatti79 said:
Is the converse true, ie cos (t) tends to 0 faster than y tends to 0.

Does cos(t) → 0 when t → 0 at all?
 
  • #8
LCKurtz said:
Does cos(t) → 0 when t → 0 at all?

Sorry, that should be cos(t) to 1 as t to 0
 
  • #9
I like Serena said:
Thanks!

You're moving on to a more complex limit.
Does that mean you got the other limits?

To find
[tex]\lim\limits_{t \to 0} {\sin t \over t}[/tex]
you should apply l'Hôpital's rule:
http://en.wikipedia.org/wiki/L'Hôpital's_rule
The article gives your limit as an example.

To answer your question, sin t and t approach 0 with equal speed.

Here is another example I am checking using different methods
[itex]\frac{1-t^2+3t^5}{t^2+4}[/itex] as t tends to 0

Method 1: Limit of the quotients is a quotient of the limits hence [itex]\frac{1-(0)^2+3(0)^5}{(0)^2+4}=\frac{1}{4}[/itex]

Method 2: [itex]\frac{f(t)}{g(t)}=\frac{f'(t)}{g'(t)}[/itex] as t tends to 0 implies [itex]\frac{15t^4-2t}{2t}[/itex] as t tends to 0 gives [itex]\frac{0}{0}[/itex]...?

Method 3: [itex]\frac{1/t^2-1/t^3+3}{1/t^3+4/t^5}=\frac{3}{0}[/itex]...?

I have seen each of these methods work correctly in other problems...whats going on?
 
  • #10
bugatti79 said:
Here is another example I am checking using different methods
[itex]\frac{1-t^2+3t^5}{t^2+4}[/itex] as t tends to 0

bugatti79 said:
Method 1: Limit of the quotients is a quotient of the limits hence [itex]\frac{1-(0)^2+3(0)^5}{(0)^2+4}=\frac{1}{4}[/itex]

This is correct.
bugatti79 said:
Method 2: [itex]\frac{f(t)}{g(t)}=\frac{f'(t)}{g'(t)}[/itex] as t tends to 0 implies [itex]\frac{15t^4-2t}{2t}[/itex] as t tends to 0 gives [itex]\frac{0}{0}[/itex]...?

L'Hôpital's rule is only applicable if nominator and denominator both approach zero, meaning you have the indeterminate form 0/0.
This is not the case now.
bugatti79 said:
Method 3: [itex]\frac{1/t^2-1/t^3+3}{1/t^3+4/t^5}=\frac{3}{0}[/itex]...?

This form only works if t approaches infinity, which it doesn't in this case.
 
  • #11
I like Serena said:
L'Hôpital's rule is only applicable if nominator and denominator both approach zero, meaning you have the indeterminate form 0/0.
This is not the case now.

Does this work for an indeterminate form of infinity over infinity also? Thanks
 
  • #12
bugatti79 said:
Does this work for an indeterminate form of infinity over infinity also? Thanks

Actually, yes it works, as you can see in the wiki article.
 

1. What is a limit in calculus?

A limit in calculus is a fundamental concept used to describe the behavior of a function as the input values get closer and closer to a specific value. It is denoted by the symbol "lim" and is used to determine the value that a function approaches as the input values approach a certain point.

2. How do you calculate a limit?

To calculate a limit, you first need to determine the value that the input is approaching. Then, you evaluate the function at values that are closer and closer to the given point. If the values approach a specific number, that number is the limit. If the values approach different numbers from the left and right sides, then the limit does not exist.

3. What is the difference between a one-sided and two-sided limit?

A one-sided limit only considers the values of the function as the input approaches the given point from one direction, either from the left or right side. A two-sided limit considers the values of the function as the input approaches the given point from both the left and right sides.

4. When does a limit not exist?

A limit does not exist when the values of the function approach different numbers from the left and right sides, or when the function has a vertical asymptote at the given point. It also does not exist if the function oscillates or has an infinite number of discontinuities at the given point.

5. Can limits be calculated for all functions?

No, not all functions have well-defined limits. Functions that have vertical asymptotes, oscillate, or have an infinite number of discontinuities do not have limits. Additionally, some functions may have limits at some points but not at others.

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