# Limit of a ratio

1. Jan 9, 2014

### bobie

1. The problem statement, all variables and given/known data
This is not homework
Could you help me understand what happens to this function
$$\frac{1}{\sqrt{100..1^2+199...9}- 100...1}$$
when the dots are filled with 0's and 9's , shouldn't the result be 1/.99999.....)= 1.?

but if to this input (http://m.wolframalpha.com/input/?i=1/+((sqrt(100000001^2+199999999)-100000001)&x=9&y=1) ≈ 1.000 000 02,
I get: ≈ 0.00002262 which is 1/44208.7

shouldn't it be 1.000 000 002 ?

Last edited: Jan 9, 2014
2. Jan 9, 2014

### Ray Vickson

How many 0's and 9's are there to be filled in? If the 0's and 9's are never ending, then there is no such a thing as 100....^2 or 199... : these are just ∞. Also, in this case there is no such a thing as a last digit as in your 100...1 and 199...9. On the other hand, if the number of filled-in 0's and 9's is finite and limited, the answer will depend to some extent on exactly how many are used, and also whether the number of filled-in 0's is the same as the number of 9's.

3. Jan 9, 2014

### bobie

Ray, you can check that in the second link there is just one more 0,9,0, (10 digits vs 9 digits in the first,
why the result?

4. Jan 9, 2014

### DeIdeal

See the small subscript 2 on the 1000000001s of the second link? That means Wolfram interprets them as binary numbers, so you definitely won't get the results you were expecting. This is how you get approx. 1.000000002.

5. Jan 9, 2014

### D H

Staff Emeritus
bogie, the problem is how wolfram alpha parses your expressions. In your first link, wolfram alpha interprets 100000001 as a being decimal number. In your second link, it interprets 1000000001 as being a binary number.

Edit DeIdeal beat me to it!

6. Jan 9, 2014

### bobie

Thanks, I eluded the problem with an artifice (100000000000-1*-1)):http://m.wolframalpha.com/input/?i=...000000000*1-1^2)-(100000000000-1*-1))&x=9&y=8
Di you know if it is possible to get a curve at alpha , just giving four or five points?

7. Jan 9, 2014

### D H

Staff Emeritus
8. Jan 9, 2014

### Ray Vickson

The result is approximately 10.000 ... (not nearly 1.0!).Wolfram Alpha delivers this if you write something that forces it to interpret 1000000001 in decimal instead of binary. For example, if we write 1000000005-4 instead, we get correct results;
see http://www.wolframalpha.com/input/?i=1%2F%28sqrt%28%281000000005-4%29^2%2B199999999%29-%281000000005-4%29%29

It is useful to re-write the problem in a more "understandable" form. Your quantity is
$$F = \frac{1}{\sqrt{a^2 + b} - a}, \\ \text{where } a = 1000000001, \; b = 199999999$$
Since $b << a^2$ we can get a rapidly-converging series expansion that is much more insightful than the original expression. We can write
$$F = \frac{1}{a} \frac{1}{\sqrt{1+r}-1}, \; r = b/a^2.$$
For small $r$ we have a rapidly-converging expansion
$$\sqrt(1+r)-1 = \frac{1}{2}r - \frac{1}{8} r^2 + \frac{1}{16}r^3 - \frac{5}{128} r^4 + \frac{7}{256} r^5 + \cdots$$
In our case $r = b/a^2 \doteq 0.1999999986e-9,$ so each new term is about a billion times smaller than the preceding term. Just taking a few terms should give very high precision. Plugging that series for the denominator of F and converting that to another series, we finally get
$$F = 2\frac{a}{b} +\frac{1}{2}\frac{1}{a} -\frac{1}{8} \frac{b}{a^3} +\frac{1}{16} \frac{b^2}{a^5} -\frac{5}{128} \frac{b^3}{a^7} +\frac{7}{256} \frac{b^4}{a^9} + \cdots$$
For the current case the successive terms in the expansion of F are
$$\text{term 1} \doteq 10.000000060000000300\\ \text{term 2} \doteq 0.49999999950000000050e-9 \\ \text{term 3} \doteq -0.24999999800000000525e-19\\ \text{term 4} \doteq 0.24999999625000002250e-29 \\ \text{term 5} \doteq -0.31249999312500006500e-39 \\ \text{term 6} \doteq 0.43749998731250016406e-49$$
This shows very clearly that F is very near 10.0. The above input for Wolfram Alpha gives 10.000000060500000299475001500702507498937225001732106.