# Calculate the following limits Part 1

1. Nov 10, 2011

### bugatti79

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

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

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

Last edited by a moderator: Nov 10, 2011
2. Nov 10, 2011

### 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)?

3. Nov 10, 2011

### Staff: Mentor

I don't understand the question. Is the limit as t --> infinity? You have x.
If the limit is as t gets large, then yes.
What do tan-1(t) and e-2t approach as t gets large?

4. Nov 10, 2011

### Staff: Mentor

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

5. Nov 11, 2011

### bugatti79

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. Nov 11, 2011

### I like Serena

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.

7. Nov 11, 2011

### LCKurtz

Does cos(t) → 0 when t → 0 at all?

8. Nov 11, 2011

### bugatti79

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

9. Nov 11, 2011

### bugatti79

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?

10. Nov 11, 2011

### I like Serena

This is correct.

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.

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

11. Nov 11, 2011

### bugatti79

Does this work for an indeterminate form of infinity over infinity also? Thanks

12. Nov 11, 2011

### I like Serena

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

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