Question

evaluate this limit

Answer 1

\[\lim_{x \rightarrow 5}(\frac{ 1 }{ x }-\frac{ 1 }{ 5 })(\frac{ 1 }{ x^3-4x^2-5x })\]

Answer 2

0?

Answer 3

seems why you graph it it tends to -1/750

Answer 4

when*

Answer 5

I wonder... could we instead, reduce it to a single fraction bar?
So much simpler that way...

Answer 6

(5-x)/(5x*(x^3 - 4x^2 -5x)), i think it is infinity cuz the denominator is zero too when x->5

Answer 7

\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x(x^3-4x^2-5x) }\]

Answer 8

\[\lim_{x \rightarrow 5}\left(\frac{5-x}{5x}\right)\left(\frac{ 1 }{ x^3-4x^2-5x }\right)\]
\[\lim_{x \rightarrow 5}\left(\frac{5-x}{5x( x^3-4x^2-5x)}\right)\]

\[\Large\lim_{x \rightarrow 5}\left(\frac{5-x}{5 x^4-20x^3-25x^2}\right)\]

Answer 9

seperate now?

Answer 10

Are we, perhaps, allowed to use L'Hôpital?

Answer 11

nup!

Answer 12

says not to

Answer 13

ohh.. Well, no problem :D

Answer 14

wait it doesn't say u need to or not. thats for the other question

Answer 15

We just have to identify the largest exponent which x is given, in this case, 4
\[\Large\lim_{x \rightarrow 5}\left(\frac{5-x}{5\color{red}{ x^4}-20x^3-25x^2}\right)\]
right?

Answer 16

now u divide everything by x^4?

Answer 17

Let's be formal... we multiply everything by...
\[\LARGE \frac{\frac1{x^4}}{\frac1{x^4}}\]

Answer 18

oh k fair.

Answer 19

i get 0?

Answer 20

On second thought, that may not work out as intended... recalculating :3

Answer 21

I forgot it wasn't a limit to infinity... sorry, old habits die hard :D

Answer 22

haha

Answer 23

Start from here...
\[\LARGE \lim_{x \rightarrow 5}\quad\frac{ 5-x }{ 5x(x^3-4x^2-5x) }\]

Answer 24

And factor the denominator.

Answer 25

If they are both zero at x = 5, then that must mean x-5 is a factor of the denominator.

Answer 26

\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x(x-5)(x)(x+1) }\]

Answer 27

I bet you can see where this is going now? :)

Answer 28

\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x^2(x-5)(x+1) }\]

Answer 29

YEP

Answer 30

i can just put a negaitve and change it to x-5

Answer 31

Yeah, sorry for that false lead :)
Let me just make that tweak, to, err, make amends XD
\[\lim_{x \rightarrow 5}\frac{ -1(x-5) }{ 5x^2(x-5)(x+1) }\]

Answer 32

haha thats why graphing is good! yep it comes to -1/750

Answer 33

gooood!

Answer 34

what about this one!

Answer 35

\[\lim_{x \rightarrow 0}\frac{ \tan4x }{ 5x^2+2x }\]

Answer 36

no L'Hopital though

Answer 37

Ooooh... let's have a look...

Answer 38

we can factor out x?

Answer 39

then we have sin4x/x i think i know where this might go to!

Answer 40

okay... tell me...

Answer 41

gimmi a sec

Answer 42

\[\lim_{x \rightarrow 0}\frac{ \sin4x }{ x }\frac{ 1 }{ \cos4x(5x+2) }\]

Answer 43

now we multiply both numerator and denominator by 4

Answer 44

Just go on...

Answer 45

While I silently ponder whether or not you needed my help with this question at all XD

Answer 46

\[\lim_{x \rightarrow 0}\frac{ \sin4x }{ 4x }\frac{ 4 }{ \cos4x(5x+2) }\]
\[\lim_{4x \rightarrow 0}\frac{ \sin4x }{ 4x }\lim_{x \rightarrow 0}\frac{ 4 }{ \cos4x(5x+2) }\]

Answer 47

Of course, that second line is only possible because both limits exist ;)

Answer 48

And so the limit of the entire thingy is...?

Answer 49

2!

Answer 50

ah yes! since both limits exist. thats important

Answer 51

And... you're right.
So... remind me again what you needed me for? ^_^

Answer 52

hahaha re-assurance!

Answer 53

and ur crazy good!

Answer 54

except ucouldn't help me with that linear algebra proof :( i think i nutted it out though not sure

Answer 55

I don't really do linear algebra... matrices hate me ^_^

Answer 56

haha ohh well! thanks for that! terenz to the rescue once again!

Answer 57

^_^