Question

some amazing questions...

Answer 1

yaaay :3

Answer 2

\[\Large \lim_{x \rightarrow 0} x \left[ \frac{ 1 }{ x } \right] \]

\[ \Large \lim_{x \rightarrow 0} \frac{ \left[ x \right] }{ x } \]

find \(a\) and \(b\) which \( \large \lim_{x \rightarrow 0} \frac{ \sqrt{ax +b}-2 }{ x } = 1 \)

i've already solved all of them,i'm sure i have solved the third one correctly but i'm not sure of the first and second one.

Answer 3

let me write the solution of the third one...

Answer 4

Do the [square brackets] mean something special?

Answer 5

[] = \(\lfloor \rfloor \)

floor function

Answer 6

yes means least integer

Answer 7

[x] gives the greatest integer less than or equal to x

Answer 8

yes

Answer 9

guys can i post the solution for the third one?

Answer 10

post im waiting :O

Answer 11

if we suppose \(\large \sqrt{ax + b} = t\) so if \(x \rightarrow 0 \) then \(t \rightarrow \sqrt{b} \) and \(\large x = \frac{ t^2 - b }{ a }\)

so,
\[\large\lim_{x \rightarrow 0} \frac{ \sqrt{ax + b}-2 }{ x } = \lim_{t \rightarrow \sqrt{b}} \frac{ (t-2)a }{ (t -\sqrt{b})(t+\sqrt{b}) }\]

if we suppose \( ~ t-\sqrt{b} = t-2 ~ \color{red}{(1)}\) then \(\lim_{t \rightarrow \sqrt{b}} \frac{ (t-2)a }{ (t -\sqrt{b})(t+\sqrt{b}) } = \lim_{t \rightarrow \sqrt{b}} \frac{ a }{ t+\sqrt{b} } = 1\)

so,\(\large \frac{ a }{ 2\sqrt{b} } = 1\) and from the equation \(\large \color{red}{(1)} \) we know \(b=4\)

so \(\Large a=4 \) and \(\Large b=4 \).

Answer 12

and for first one i supposed that when \(x \rightarrow 0\) then \(\left[ \frac{ 1 }{ x } \right] \rightarrow 0\)
so \( \large \lim_{x \rightarrow 0} ~ x \left[ \frac{ 1 }{ x } \right] = 0 \)

Answer 13

but i'm not sure whether i've used properties correctly

Answer 14

\[\lim \limits_{x\to 0} x\left[ \frac{1}{x}\right] = \lim \limits_{y\to \infty } \frac{\left[y\right]}{y} = \lim \limits_{y\to \infty } \frac{y - \left\{ y\right\}}{y} = \lim \limits_{y\to \infty } 1-\frac{\left\{ y\right\}}{y} = 1-0 = 1\]

Answer 15

when [x] =x (means x is integer ) then limit is 1 at x tends to zero
otherwise
[x] <x means x have radical part then limit >1
so as x tend to [x] we would have a limit of 1

Answer 16

\(\{y\}\) is the fractional part of real number \(y\)

Answer 17

i got that

Answer 18

second limit looks very tricky

Answer 19

part b is different i guess pick x<0
[x] <x any where thus<1
if u picked small values u would got (|x0|)
0 each time
i would say limit tend to 0

Answer 20

well @ganeshie8 , would you explain ut solution? u supposed \[\frac{ 1 }{ x } = y\] ?

Answer 21

@ganeshie8

Answer 22

yes let \(\large y = \frac{1}{x}\)
as \(x\) tends to \(0\), the value of \(y\) tends to \(\infty \), yes ?

Answer 23

yes

Answer 24

wait,i got it

Answer 25

why my explanation was incorrect ?

Answer 26

as \(\large x \to 0\)

we have \(\left[\dfrac{1}{x}\right] \to \infty\)

Answer 27

wait a moment! \(y \rightarrow \infty\) then \(\frac{ \left\{ y \right\} }{ y } \rightarrow ?\)

Answer 28

Notice that the fractional part is always a number between 0 and 1

Answer 29

correct , so \(\left\{ y \right\}\) would be smaller and smaller and \(y\) would be bigger so the whole fraction would be 0

Answer 30

\[\large 0 \le \{y\} \lt 1\]
\[\large \frac{0}{y} \le \frac{\{y\}}{y} \lt \frac{1}{y}\]
\[\large 0 \le \frac{\{y\}}{y} \lt \frac{1}{y}\]

now take limit through out
\[\large \lim \limits_{y\to \infty}0 \le \lim \limits_{y\to \infty} \frac{\{y\}}{y} \lt \lim \limits_{y\to \infty}\frac{1}{y}\]

\[\large 0 \le \lim \limits_{y\to \infty} \frac{\{y\}}{y} \lt 0\]

Answer 31

so that value of that limit is between 0 and 0
that means the value is exactly 0

Answer 32

@ganeshie8 can i use squeeze theorem for the last line u wrote?

Answer 33

yes we are using squeeze thm above ^

Answer 34

got it ;) and why my explanation was incorrect?

Answer 35

why is it the case that as \(\large x\to 0\), the value of \(\large \left[\dfrac{1}{x}\right]\to 0\) ??

doesn't that look wrong at its face value ?

Answer 36

I am refering to this
http://gyazo.com/03ca984e638b6eafd5975a773199b404

Answer 37

that page doesn't load for me.
yes...
and ur idea about the third one ?

Answer 38

im still going through it one sec

Answer 39

Wow! it took me a while to see why it works xD
it looks brilliant!! it works

Answer 40

im talking about limit #3

Answer 41

ok lets note something
\(\left[ 1/x\right]\neq1/ \left[ x\right]\)

Answer 42

thus i guess explanation was true !

Answer 43

yes your solution for limit #3

Answer 44

no mistakes
i have alternative method using definition of derivative which il post in the end maybe.. but ur method looks perfect !

Answer 45

so looks like we are done with first and third limits ?

Answer 46

oh thank you;) but i would like to now how u used derivative definition for it

Answer 47

post it later

Answer 48

yes only the second one is left...i think @Marki has some ideas for it

Answer 49

ur idea's were so cool xD
such that i cant think of anything

Answer 50

im making piecewise function :3

Answer 51

so around zero