Question

coudl you guys help me evalating this limit, only using algebra techniques.

Answer 1

the textbook answer seems to be sqrt(6/5) or 1/5sqrt(30) but i have no idea how to get there tried factoring both sides but no success yet

Answer 2

As x goes to -3? Or as y goes to -3? (You're function only has y's in it. If the limit is, indeed, as x goes to -3, this is really easy).

Answer 3

omg my mistake

Answer 4

yeh a x goes to -3 replace y for x :D

Answer 5

No worries. I just wanted to make sure it wasn't your teacher trying to be tricky!

Using only algebra...Hmm.

Answer 6

yeah im still not allowed to use L'hospital rule

Answer 7

Factor the equation under the sq root.

Answer 8

yeah i factored top like (x-3)(x+3)

Answer 9

did you try factoring top and bottom yet?

Answer 10

Now do the bottom

Answer 11

The denominator factors nicely, as well. A hint will be that we want something on the top to cancel with something on the bottom.

Answer 12

and bottom using quadratic formula into (x+3)(x+1/2)

Answer 13

but the answer doesnt match with the graph or the textbook answer :(

Answer 14

No that is not the bottom

Answer 15

i did it using quadratic formula

Answer 16

x+1/2=0
2x+1=0
2x+1 should be the other factor

Answer 17

oh how you got that quadratic :D

Answer 18

Well I know (x+3)(x+1/2) won't work because the leading coefficient will be x^2 not 2x^2

Answer 19

and you said you got x=-1/2
multiply 2 on both sides
2x=-1
add 1 on both sides
2x+1=0
so 2x+1 is a factor

Answer 20

\(\bf \textit{quadratic formula}\\
y={\color{blue}{ 2}}x^2{\color{red}{ +7}}x{\color{green}{ +3}}
\qquad \qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

Answer 21

leading term will be x^2 not 2x^2*

Answer 22

That isn't really the normal way I factor but yeah you can work backwards if you work backwards correctly to get the factors

Answer 23

\[\sqrt{\frac{ (x+3)(x-3) }{ (2x+)(x+3)}}\]

Answer 24

So Bryan what do you do next?

Answer 25

hmmm retry the quadratic... though... plain factoring works well though.. anyhow as nelsonjedi pointed out already

\(\bf lim_{x\to -3}\ \sqrt{\cfrac{x^2-9}{2x^2+7x+3}}\quad
\begin{cases}
x^2-9\to (x-3)(x+3)\\
2x^2+7x+3\to (2x+1)(x+3)
\end{cases}\quad thus
\\ \quad \\
\sqrt{\cfrac{(x-3)\cancel{ (x+3) }}{(2x+1)\cancel{ (x+3) }}}\)

Answer 26

so i tried quadratic but still the same result could you guys point out my mistake

Answer 27

so it cant be done with quadratic equation?

Answer 28

FOIL method then?

Answer 29

FOIL...that's the ticket

Answer 30

oh isee then Thank you for all the help !

Answer 31

but it really hard i see

Answer 32

Do yiu know thw answer yet?

Answer 33

oh got it

Answer 34

hahaha not hard really i jsut had forgotten that trick

Answer 35

You cannot use quadratic to factor an equation..I know I have made the mistake in the past. Quadratic formula is a method to solve an equation...

Answer 36

So what is the limit?

Answer 37

thank for the advise i was totally sure that worked if i set the thing to 0

Answer 38

it appears to be -6 over -6

Answer 39

Zero is correct

Answer 40

whoa but i doesnt match the answer on the book

Answer 41

are you guys still on the same problem?

Answer 42

I will copy what @jdoe0001 said above \[\bf lim_{x\to -3}\ \sqrt{\cfrac{x^2-9}{2x^2+7x+3}}\quad
\begin{cases}
x^2-9\to (x-3)(x+3)\\
2x^2+7x+3\to (2x+1)(x+3)
\end{cases}\quad thus
\\ \quad \\
\sqrt{\cfrac{(x-3)\cancel{ (x+3) }}{(2x+1)\cancel{ (x+3) }}}\]

Answer 43

if you put in -3 what do you get?

Answer 44

i got -6/-6

Answer 45

not exactly

Answer 46

good on the top

Answer 47

not on the bottom

Answer 48

2(-3)+1=-6+1=-5

Answer 49

omg

Answer 50

lol such a silly mistake

Answer 51

hahah thanks for pointing it out

Answer 52

\[\sqrt{\frac{-3-3}{2(-3)+1}}=\sqrt{\frac{-6}{-6+1}}=\sqrt{\frac{-6}{-5}}\]

Answer 53

:D Thank You

Answer 54

and then of course that inside simpliflies to 6/5

Answer 55

yeah now thats the answer