Question

Lim of expression below as x approaches -4.

Answer 1

\[\frac{ \sqrt{x^2-9}-5 }{ x+4 }\]

Answer 2

Why is that tricky? Denominator approaches zero numerator doesn't. This puppy blows up, no?

Answer 3

man... i didn't even try to plug in 4 i just assumed i had to multiply by conjugate so i did and went on an endless expression... tthanks

Answer 4

oh wait nevermind

Answer 5

it is +9

Answer 6

Do NOT "plug in" anything, every. Give it some thought.

Approaches zero.
Approaches something finite.

Absolutely do NOT substitute anything unless you know it is continuous. This isn't.

Answer 7

No, there is not a finite limit. It increases (or decreases) without bound.

Answer 8

Tkhunny you have to multiply by the conjugate.

Answer 9

no i'm saying it should be x^2 +9

Answer 10

typo

Answer 11

Let me type out the answer.

Answer 12

Now you don't. If you want more information, maybe that is so. To answer the problem, as stated, ain't nobody got time for that.

Answer 13

ok so i did conjugate and got \[\frac{ x^2-16 }{ x \sqrt{x^2+9} +5x +4\sqrt{x^2+9}+20}\]

Answer 14

tkhunny the expression i wrote had a typo, it is X^2+9 not - 9

Answer 15

Well, why didn't you say so?

Answer 16

from there i tried L.H rule and got something and plugged in -4 and got -.2915 but that is wrong

Answer 17

haha I tried

Answer 18

Let's start by multiplying the conjugate. We see that the root in the numerator as a - so we need to multiply by the conjugate.
Let's start by multiplying the numerator by root(x^2 +9) +5 (This is the conjugate noticed how it is now +5).
When we multiply the roots cancel and the 5 and -5 multiply to -25.
We get: x^2 +9 -25= x^2 -16
Now the denominator: We can simply say (x+4)(root(x^2 +9) +5)

But look! Our numerator can factor to (x-4) and (x+4). Now our x-4 in the numerator and denominator can cancel leaving us: x+4/(root(x^2 +9)+5)
All we do now is simply plug in -4.
We will get 0 in the numerator and 10 in the denom. however ultimately making the answer 0.

Answer 19

That's the answer. I am in calc but we haven't learned L. hop's rule but simply using the laws of limits at a constant we can get that.

Answer 20

oh factor the numerator I see

Answer 21

ok well i put 0 in for the answer and it is wrong

Answer 22

Yo i just saw my mistake

Answer 23

Can you re-enter it?

Answer 24

lol no but i still want to figure it out.

Answer 25

Again, "plug in" doesn't mean anything. Never do it.

What we have is this other expression, \(\dfrac{x-4}{\sqrt{x^{2}+9}+5}\), which is continuous at x = -4 and is equivalent to the original expression everywhere EXCEPT x = -4. Therefore, evaluating that last expression at x = -4, will provide the limit of the original expression as x approaches -4.

Answer 26

I canceled x-4 when the x+4 canceled making it -4-4 in the numerator which is -8. The denomater when plugging in -4 will =10 making the answer -4/5. Tell me if you need to see how I got 10 in the denominator.

Answer 27

No tkhunny. Stop. you do plug in after it is safe to. It's simply algebraic limits he doesn't need to look at continuity.

Answer 28

I dare you to find me a definition of "plug in". It doesn't exist.

Answer 29

Ugh tkhunny what have u taken?

Answer 30

haha "Does not exist" get it

Answer 31

I mean are you in advanced mathematics? cause idk whether to argue or no

Answer 32

well i now see how it is -4/5 thanks guys lolo

Answer 33

Let's just say that I learned a long time ago that careless language only confuses - never helps.

There is a mathematical concept known as "substitution".
There is no such thing called "plug in".