# 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

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Homework Helper
MHB
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)?

Mentor

## 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?

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

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

bugatti79
For this limit, use [ itex] \lim_{t \to \infty} ... [ /itex ]

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.

Homework Helper
MHB

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.

Homework Helper
Gold Member
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?

Homework Helper
MHB
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

Homework Helper
MHB
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.