Calculate the following limits Part 1

[bugatti79]

Homework Statement

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

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

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

Thanks

 

[I like Serena]

Hi bugatti79!

What is the range of \tan^{-1}?
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 e^{-2t}.
What is the smallest value that it can take (but just can't reach)?

 

[Mark44]

Homework Statement

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

I don't understand the question. Is the limit as t --> infinity? You have x.

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

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?

 

[Mark44]

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

For this limit, use [ itex] \lim_{t \to \infty} ... [ /itex ]

 

[bugatti79]

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.

 

[I like Serena]

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
\lim\limits_{t \to 0} {\sin t \over t}
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.

 

[LCKurtz]

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?

 

[bugatti79]

Does cos(t) → 0 when t → 0 at all?

Sorry, that should be cos(t) to 1 as t to 0

 

[bugatti79]

Thanks!

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

To find
\lim\limits_{t \to 0} {\sin t \over t}
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
\frac{1-t^2+3t^5}{t^2+4} as t tends to 0

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

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

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

I have seen each of these methods work correctly in other problems...whats going on?

 

[I like Serena]

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

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

This is correct.

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

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

This form only works if t approaches infinity, which it doesn't in this case.

 

[bugatti79]

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

 

[I like Serena]

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.