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If #f(9)=9# and #f^'(9)=4#, then what is the value of #lim_(x to9)(sqrt(f(x))-3)/(sqrtx -3)?#
1 Answer
Explanation:
Reading speed test. We know that,
#(1)f'(c)=lim_(xtoc) (f(x)-f(c))/(x-c) ,# take #x=9#
#=>f'(9)=lim_(xto9) (f(x)-f(9))/(x-9) ,where, f(9)=9#
Markdown 3 14 – efficient markdown editor. #=>f'(9)=lim_(xto9)(f(x)-9)/(x-9) ,where, f'(9)=4#
#=>color(brown)(lim_(xto9)(f(x)-9)/(x-9)=4..(2)#
We take,
#L=lim_(xto9) (sqrtf(x)-3)/(sqrtx-3)#
#color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red)((1))xxcolor(blue)((1))#
#color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red)((sqrtf(x)+3)/(sqrtf(x)+3))xxcolor(blue)((sqrtx+3)/(sqrtx+3))#
#color(white)(L)=lim_(xto9)((sqrtf(x))^2-(3)^2)/((sqrtx)^2-(3)^2)xx(color(blue)(sqrtx+3))/(color(red)(sqrtf(x)+3))# News explorer 1 1 download free.
#color(white)(L)=color(brown)(lim_(xto9)(f(x)-9)/(x-9))xxlim_(xto9)(sqrtx+3)/(sqrtf(x)+3).tocolor(brown)([use (2)]# Postbox 6 1 12 oz.
#color(white)(L)=color(brown)(4)xx(sqrt9+3)/(sqrtf(9)+3)#
#color(white)(L)=4[(3+3)/(sqrt9+3)]#
#:.L=4{6/6}=4#
Related questions
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Main page for digit sum arithmetic
| The Equivalence of Digit Sum Arithmetic withModular 9 Arithmetic |
Modular 9 arithmetic is the arithmetic of the remainders after division by 9.For example, the remainder for 12 after division by 9 is 3. This is expressed as
12 = 3 (mod 9)
Likewise 25=7 (mod 9) and 9=0 (mod 9). Note that 12+25=37 and that 37=1 (mod 9).But 3+7=10=1 (mod 9) so the equivalent of the sum of two numbers modulo 9 isequal to the modulo 9 equivalent of the sum of their modulo 9 equivalents.
The modulo 9 equivalent of 12 is 3 which is also the digit sum of 12. This is no coincidence. There is a very close relationship between the modulo 9 equivalentsof numbers and their digit sums.
| Integer | Digit Sum | Remainder Function Mod 9 | |||||||||||||||
| Multiplication Table for Modulo 9 Arithmetic | |||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||||||||
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |||||||||
| 0 | 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 | |||||||||
| 0 | 3 | 6 | 0 | 3 | 6 | 0 | 3 | 6 | |||||||||
| 0 | 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 | |||||||||
| 0 | 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 | |||||||||
| 0 | 6 | 3 | 0 | 6 | 3 | 0 | 6 | 3 | |||||||||
| 0 | 7 | 5 | 3 | 1 | 8 | 6 | 4 | 2 | |||||||||
| 0 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
If the 0's were replaced by 9's and the table rearranged so the first column becomes the last column and the first rowbecomes the last row the result would be identical to the table for the sequences of digit sums.
| The Rearrangement of the Multiplication Table for Modulo 9 Arithmetic With 9 Substituted for 0 | ||||||||
|---|---|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 | 9 |
| 3 | 6 | 9 | 3 | 6 | 9 | 3 | 6 | 9 |
| 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 | 9 |
| 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 | 9 |
| 6 | 3 | 9 | 6 | 3 | 9 | 6 | 3 | 9 |
| 7 | 5 | 3 | 1 | 8 | 6 | 4 | 2 | 9 |
| 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 9 |
| 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 |
This table is identical with the table of digit sums for multiples of numbers. The conclusion is that digit sum arithmetic is the virtually the same as modular 9arithmetic except there is a replacement of 0's with 9's. The equivalence of 9 and 0 takes care of a smallproblem. The digit sum of all multiples of 9 is 9 except for the case of 0 times 9 which has a digit sumof 0. So the digit sum of all multiples of 9 is equivalent to 9.
In mathematical terminology thereis an isomorphism between digit sum arithmetic and modular 9 arithmetic. All ofthe properties of associativity, distributivity, communtivity, identities andadditive inverses carry over from modular 9 arithmentic to digit sum arithmetic. Multiplicativeinverses exist for some elements but not all so neither entity is a mathematical field.In the table below the cases of products that yield the multiplicative identity; i.e., 1; are shown in red.
Aol 9.5
| Multiplication Table for Modulo 9 Arithmetic | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 2 | 0 | 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 |
| 3 | 0 | 3 | 6 | 0 | 3 | 6 | 0 | 3 | 6 |
| 4 | 0 | 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 |
| 5 | 0 | 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 |
| 6 | 0 | 6 | 3 | 0 | 6 | 3 | 0 | 6 | 3 |
| 7 | 0 | 7 | 5 | 3 | 1 | 8 | 6 | 4 | 2 |
| 8 | 0 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
Goodtask 4 9 5 X 9 5
As can be seen from the above table inverses exist for 1, 2, 4, 5, 7 and 8 but not for 0, 3 and 6.
