Question

limx→-3 (√2x+22)-4/x+3

Answer 1

\(\LARGE\color{blue}{ \lim_{x \rightarrow -3} ~~\frac{\sqrt{2x+22}-4}{x+3} }\)

Answer 2

yes ?

Answer 3

yes

Answer 4

What have you tried to do ? (On this site we need to provide guidance not answers)

Answer 5

It is not the same on both sides, so the limit exists or not ?

Answer 6

exists? i think

Answer 7

No it does not.. perhaps it is +4 in your question (not -4) ?

Answer 8

so how does it not exist? how am i supposed to show work for it haha

Answer 9

\[\lim_{x \rightarrow -3}\frac{\sqrt{2x+22}-4}{x+3}\cdot\frac{\sqrt{2x+22}+4}{\sqrt{2x+22}+4}\\
\lim_{x \rightarrow -3}\frac{2x+22-16}{(x+3)(\sqrt{2x+22}-4)}\\
\lim_{x \rightarrow -3}\frac{2x+6}{(x+3)(\sqrt{2x+22}-4)}\\
\lim_{x \rightarrow -3}\frac{2(x+3)}{(x+3)(\sqrt{2x+22}-4)}\\
\lim_{x \rightarrow -3}\frac{2}{\sqrt{2x+22}-4}\]

Answer 10

But it is not equal for left and right side? 2 sided limit does not exist.

Answer 11

With +4 on the top, instead of -4, it would okay though.

Answer 12

√(2x+22)-4 → √(2{-3}+22)-4 → √(-6+22)-4 → √16 - 4

denominator = 0 or -8

Answer 13

oops :)

Answer 14

http://www.wolframalpha.com/input/?i=Limit%5B%28Sqrt%5B2x%2B22%5D-4%29%2F%28x%2B3%29%2Cx-%3E-3%5D

Answer 15

I would like to know how though

Answer 16

I think you're using a slightly incorrect conjugate? We're given - in the numerator, so you multiply by +.

Answer 17

I am not using a conjugate, I am continuing from where you left off (calculations in black latex) and just plugging -3 for x.

Answer 18

Oh I see the mistake, *I* was using the wrong conjugate :P

The last line should be
\[\lim_{x\to-3}\frac{2}{\sqrt{2x+22}~{\color{red}{\Huge{+}}}~4}\]
Thanks for catching that.

Answer 19

yes, but for √16 = -4 as well as +4, you still get
denominator = 0 and 8 (instead of -8)

Answer 20

The highest math course I took was trig, so I don't really get how is it 1/4. Wolfrrrrram

Answer 21

\[x^2=16~~\Rightarrow~~x=\pm\sqrt{16}=\pm4\]
but we're already given the positive square root, \(\sqrt{16}=4\).

Answer 22

Wolfram says the same: http://www.wolframalpha.com/input/?i=Sqrt%5B16%5D%3D-4

Answer 23

yes, so with the positive root... I agree

Answer 24

Do you always take the positive root, when solving such limits ?

Answer 25

Well, if the expression contains \(\sqrt{}\) alone (without a sign in front) we assume the positive root. The root is only negative if there's a minus sign in front.

Compare:
\[\sqrt {16}+1=4+1=5\\
1-\sqrt {9}=1+(-3)=1-3=-2\]

Answer 26

Yes... ty! I'll definitely remember that, I'll need it in calculus.

Answer 27

No problem. Good luck

Answer 28

:)