Question

@Kreshnik an easy one for you

\(\LARGE \lim_{x \rightarrow 8} \frac{16 - x}{x}\)

Answer 1

oh man i screwed up >.<

Answer 2

\[\Huge \lim_{x\to 8}\frac{16-x}{x}\]
warm up? lol

Answer 3

lol

Answer 4

ill just post the real one next time

Answer 5

nahh... it's just about substituting.. ...
\[\LARGE \lim_{x\to 8}\frac{16-x}{x}=\frac{16-8}{8}=\frac88=1\]
... ok my turn? :) (level 1 too :P )

Answer 6

\[\LARGE \lim_{x\to 3}\frac{x^2-9}{x+3}\]

Answer 7

this can be solve two ways but to avoid all the latex ill go with

3^2 - 9/ 4+3 = 9 - 9/6 = 0/6 = 0

Answer 8

that's correct ...3^2 - 9/ 4+3 i guess here yo meant 3^2 - 9/ 3+3 ;) typo... Your turn ;)

Answer 9

1 question... am I allowed to use paper? (to calculate on my notebook)

Answer 10

why not hehe

Answer 11

??

Answer 12

this is hard hmmm

Answer 13

please dont... lol hahaha...

Answer 14

\(\Large \lim_{h \rightarrow 0} \frac{(x+ h)^2 - 2(x+h) + 1 - x^2 + 2x - 1}{h}\)

think you can handle that? :D

Answer 15

multiple choices?? (I've never done such thing like that... I'll try though )

Answer 16

lol it's where derivatives come from...

\(\LARGE \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\)

but it's easy..lemme give some choices

Answer 17

is it 2x-2

Answer 18

haha you got it :DDD

Answer 19

\[\Large \lim_{h\to0}\frac{(x+h)^2-2(x+h)+1-x^2+2x-1}{h}\]
\[\Large \lim_{h\to0}\frac{x^2+2xh+h^2-2x -2h-x^2+2x }{h}\]
\[\Large \lim_{h\to0}\frac{\cancel x^2+2xh+h^2-\cancel {2x} -2h-\cancel x^2+\cancel {2x} }{h}\]
\[\Large \lim_{h\to0}\frac{h(2x+h -2) }{h}\]
2x+0-2
2x-2
:)

Answer 20

<pause> im gonna eat..

Answer 21

ok.. boun apetit ;)

Answer 22

when you're back, let me know. :)

Answer 23

im back

Answer 24

ok... should I continue with a bit harder limit or still warming up ? :D

Answer 25

you can go serious now :p

Answer 26

\[\Huge \lim_{x\to 1}\frac{\sqrt[4]{x}-1}{\sqrt[3]{x}-1}\]
A) 0 B) 2/3 C) 3/4 D) 4/3
Good Luck..

Answer 27

huh..you outdid yourself this time :D

Answer 28

can you handle that? :D.. need hints just tell me :D

Answer 29

how do you mean, "outdid" ... can you explain it.

Answer 30

would you like me to change the question? (I didn't copy this on book, it was on my mind.) next one is too... :D

Answer 31

it's a good one

Answer 32

but it's a long one too.. !! lol

Answer 33

need hints? :D

Answer 34

im thinking difference of two squares and difference of cubes but it doesnt make sense

Answer 35

substitute"
\[\LARGE a=\sqrt[3]{x} \quad \quad and\quad \quad b=1\]
and you have:
\[\LARGE a^3-b^3=(a-b)(a^2+ab+b^2 )\]
so in your problem you have a-b
so you need to multiply by
a^2+ab+b^2 to get difference of cubes...
(you'll have to do the same with numerator too. !) so they would cancel out!

Answer 36

anyway, USE L'Hopital and tell me what do you get, so we can skip this boring limit :D

Answer 37

USE L'Hopital...

Answer 38

i thought you wont let me use it :P

Answer 39

do it

Answer 40

lol how many times am i gonna lhopital this @_@

Answer 41

not much... O_O show what you got:



\[\LARGE \begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\sqrt[4]{x} - 1}}{{\sqrt[3]{x} - 1}}} \right)' \\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\frac{1}{{4\sqrt[4]{{{x^3}}}}}}}{{\frac{1}{{3\sqrt[3]{{{x^2}}}}}}}} \right)\\ = \left( {\frac{{\frac{1}{{4\sqrt[4]{{{1^3}}}}}}}{{\frac{1}{{3\sqrt[3]{{{1^2}}}}}}}} \right)\\ = \left( {\frac{{\frac{1}{4}}}{{\frac{1}{3}}}} \right) = \frac{3}{4}\end{array}\]

Answer 42

:D ... YOUR TURN. ...

Answer 43

YOU WIN! lol you made me feel incapable hahaha

Answer 44

NO NO ... there's a thousand limits that we can get...Give me a real tough one, ... I always admited that my math sucks... but i want to learn ... don't feel bad for making a mistake which has not allowed you to get the answer. I had many of those, I'll have now I'm sure lol.... Go ahead your turn. ;)

Answer 45

here's how it goes WITHOUT L'HOPITAL lol (now you'll know why I let you use L'Hopital)... I didn't want you to pass all through this. ;) (I wrote this on my programm while we were talking... )

isn't it Cute?

\[\Large \begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\sqrt[4]{x} - 1}}{{\sqrt[3]{x} - 1}}} \right)\\\\\sqrt[4]{x} = a\\1 = b\\{a^4} - {b^4} = \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\\{a^4} - {b^4} = \left( {a - b} \right)\left( {{a^3} + a{b^2} + b{a^2} + {b^3}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{a - b}}{{\sqrt[3]{x} - 1}} \cdot \frac{{{a^3} + a{b^2} + b{a^2} + {b^3}}}{{{a^3} + a{b^2} + b{a^2} + {b^3}}}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{a^4} - {b^4}}}{{\sqrt[3]{x} - 1\left( {{a^3} + a{b^2} + b{a^2} + {b^3}} \right)}}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{{\left( {\sqrt[4]{x}} \right)}^4} - {1^4}}}{{\sqrt[3]{x} - 1\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right)\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1}}{{\sqrt[3]{x} - 1\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right)\\\\now\\\sqrt[3]{x} = a\\1 = b\\{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1}}{{a - b\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right) \cdot \left( {\frac{{\left( {{a^2} + ab + {b^2}} \right)}}{{\left( {{a^2} + ab + {b^2}} \right)}}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1\left( {{a^2} + ab + {b^2}} \right)}}{{{a^3} - {b^3}\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right)\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1\left( {{{\left( {\sqrt[3]{x}} \right)}^2} + \sqrt[3]{x} + {1^2}} \right)}}{{{{\left( {\sqrt[3]{x}} \right)}^3} - {1^3}\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right)\\\\\mathop {\lim }\limits_{x \to 1} \left( {\frac{{x - 1\left( {{{\left( {\sqrt[3]{x}} \right)}^2} + \sqrt[3]{x} + {1^2}} \right)}}{{x - 1\left( {{{\left( {\sqrt[4]{x}} \right)}^3} + \sqrt[4]{x} + {{\left( {\sqrt[4]{x}} \right)}^2} + 1} \right)}}} \right)\\\\ = \left( {\frac{{\left( {{{\left( {\sqrt[3]{1}} \right)}^2} + \sqrt[3]{1} + {1^2}} \right)}}{{\left( {{{\left( {\sqrt[4]{1}} \right)}^3} + \sqrt[4]{1} + {{\left( {\sqrt[4]{1}} \right)}^2} + 1} \right)}}} \right)\\\\ = \left( {\frac{{1 + 1 + 1}}{{1 + 1 + 1 + 1}}} \right) = \frac{3}{4}\end{array}\]