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Calculate the following limits Part 1

  • Thread starter bugatti79
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1. The problem statement, all variables and given/known data

[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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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)?
 
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1. The problem statement, all variables and given/known data

[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.
First term = [itex]=\frac{1/t^2+1}{1/t^2-1}=-1[/itex]
If the limit is as t gets large, then yes.
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?
 
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[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 ]
 
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.
 

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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.
 
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?
 

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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]
This is correct.


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.


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.
 
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
 

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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.
 

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