Question

Help me find a limit! Will fan and medal!

Answer 1

lim x-> -6 for (sqrt(x^2 + 12x + 36))/ (x+6)

Answer 2

This is a vertical limit I am looking for, correct?

Answer 3

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~6}\frac{\sqrt{x^2+12x+36}}{x+6}}\)

can you factor inside the root ?

Answer 4

oh, my bad, it is \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}\frac{\sqrt{x^2+12x+36}}{x+6}}\), but still factor inside the root.

Answer 5

so it would then be sqrt ((x+6)^2)?

Answer 6

and then x+6/x+6 as a whole

Answer 7

yes, and sqrt((x+6)^2) is ?

Answer 8

\(\large\color{slate}{\displaystyle\sqrt{A^2}=A}\) you know that correct?

Answer 9

Yes. I know it would simplify to (x+6)/(x+6). would that then be one or zero?

Answer 10

I am seeing that these could cancel each other out to be 1/1 or if I were to first fill it in with x=-6, it would be zero? But neither of these seem to make sense with the graph?

Answer 11

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}\frac{\sqrt{x^2+12x+36}}{x+6}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}\frac{\sqrt{(x+6)^2}}{x+6}}\)

Answer 12

don't look at the graph (or, not yet at least).

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}\frac{\sqrt{(x+6)^2}}{x+6}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}\frac{x+6}{x+6}}\)


\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6} 1}\) good.

Answer 13

when you have a limit of a constant, it is just that constant. (because there is no x to plug in for)

Answer 14

So, the limit is -6.

Answer 15

you probably tried looking at f(-6), and that is different.

Answer 16

I am confused. My class taught me to then sub -6 in for x because it was asking for the limit of this graph as x approaches -6?

Answer 17

f(-6) does not exist, because there is a restriction in the initial expression for x to equal -6 (zero in denominator).

Now, you general find restrictions before simplifying, so there is on \(\large\color{slate}{\displaystyle x\ne -6 }\)

Answer 18

there are two parts for this limit, as it approaches 6 fro,=m the left, and as it is approaches 6 from the right.

when you find the limit, you COULD simplify, and then evaluate the limit, and that is what we did, we got:
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6} 1}\)

and there is nothing to plug in for x, so you have: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6} 1=1}\)

Answer 19

That isn't even an option given in my multiple choice? And that doesn't seem to make sense based on what I was taught?

Answer 20

@SolomonZelman I was given options a) does not exist b)36 c)0 d) infinity e) -infinity
... I understand how simplifying brought it to one, but it really is not an option.

Answer 21

it is 1, unless you wrote the limit incorrectly.

Answer 22

I did not. Are you sure I shouldn't have plugged in x=-6 before it was simplified for one to give zero? I know that there is not a vertical asymptote there. Should I look at the graph for further confirmation? There is definitely points at x=1, too, there is no vertical limit there.

Answer 23

@SolomonZelman

Answer 24

it doesn't exist I guess.

Answer 25

I didn't consider your intial function, only the way it is simplified.

Answer 26

My calculator is having a hard time giving a picture of the original function around -6... is there any chance it is infinity or negative infinity? I can't imagine it would be but would like to double check

Answer 27

But you want \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6}}\)

https://www.desmos.com/calculator/ecfwroztz0

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6+}f(x)=1}\)
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-6-}f(x)=-1}\)

Answer 28

Then it would not exist because the two sides do not agree?

Answer 29

yes

Answer 30

\[\sqrt{(x+6)^2}=x+6 \text{ if } x+6>0 \\ \sqrt{(x+6)^2}=-(x+6) \text{ if } x+6<0\]

Answer 31

wait, but here x+6=0

Answer 32

well know for limits we don't consider what happens at x=-6
we consider what happens around

Answer 33

It was for sure does not exist. Thank you so much. That was a strange problem, thank you very much for your help.

Answer 34

if x<-6 then we are consider the left limit
if x>-6 then we are consider the right limit

Answer 35

since the left and right limit do not equal the actual limit does not exist

Answer 36

yes.

Answer 37

and from the left side =-1, and right =1.

Answer 38

\[\lim_{x \rightarrow -6^-}\frac{|x+6|}{x+6}=\lim_{x \rightarrow -6^-}\frac{-(x+6)}{x+6} =-1 \\ \lim_{x \rightarrow -6^+} \frac{|x+6|}{x+6}=\lim_{x \rightarrow 6^+}\frac{x+6}{x+6}=1\]

Answer 39

yup (not like I am verifying it for you, just agreeing)

Answer 40

Ok I was just trying to explain
because there looked like there was some confusion on how to evaluate the above limit