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### Limits - Indeterminate Forms

```
Date: 10/12/97 at 13:43:29
From: Mary
Subject: Limits - Indeterminate Forms

I cannot do the problem

lim ((1/x) - (cot x))
x>0

I realize that this needs to be converted into the form 0/0 and then I
must use L'Hopital's Rule; however I have tried and I cannot do this.

I am having a similar problem with the question

lim (1/x)(ln (7x+8)/(4x+8))
x>0

Thank you.
```

```
Date: 10/12/97 at 17:45:31
From: Doctor Anthony
Subject: Re: Limits - Indeterminate Forms

>lim ((1/x) - (cot x))
>x>0

1      cos(x)
Write this as  ---  -  ------
x      sin(x)

sin(x) - x.cos(x)
-----------------
x.sin(x)

If you expand sin(x) and cos(x) in series, you get

x - x^3/3! + x^5/5! - ... - x[1 - x^2/2! + x^4/4! - ..]
-------------------------------------------------------
x[x - x^3/3! + x^5/5! - ...]

-x^3/3! + x^5/5! - ... + x^3/2! - x^5/4! + ...
-----------------------------------------------
x^2 - x^4/3! + x^6/5! - ......

divide top and bottom by x^2

-x/3! + x^3/5! - .... +  x/2! - x^3/4! + ...
--------------------------------------------
1 - x^2/3! + x^4/5! - .........

now if we let x -> 0 the top line -> 0 , but the bottom line -> 1,
and so the final result  -> 0/1   = 0

>lim (1/x)(ln (7x+8)/(4x+8))
>x>0

Again, series expansion is a good way to find the limit.

7x+8           3x
----  =  1 +  -----
4x+8          4x+8

and expand ln(1+y) as  y - y^2/2 + y^3/3 - y^4/4 + ......

3x       9x^2         27x^3
ln(1 + 3x/(4x+8)) = ------ - ---------  + ---------- - ........
4x+8     2(4x+8)^2    3(4x+8)^3

3         9x       27x^2
and so (1/x)ln(1 + 3x/(4x+8)) = ----- - --------- + --------  - ....
4x+8    2(4x+8)^2  3(4x+8)^3

Now let x -> 0 and this expression reduces to  3/8

So the limit as x -> 0 will be  3/8

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Sequences, Series

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