Question

Find the indicated limit, if it exists.

Answer 1

we need x-> (-1)

for the left sided limit, the part of the function you are using is "x-4"
for the right sided limit, the part of the function you are using is "x+6"

Answer 2

plug in -1 into the "4-x" to find the left side of the limit
plug in -1 into the "x+6" to find the right side of the limit

if these two results are equivalent to each other, then whatever the value they are both equal to, THIS, will be your answer for the limit[x-> -1] .

if those results won't be equivalent, then sate that limit DNE.

Answer 3

so since they are both equal, since they = 5, does that mean the indicated limit is 5?

Answer 4

yes@!

Answer 5

\[\lim_{x \rightarrow -1}~f(x)=5\]

Answer 6

and hence
it is continuous, because:
\[\lim_{x \rightarrow -1^-}~f(x)=\lim_{x \rightarrow -1^+}~f(x)=5=f(5)~~~~~~~~=5\]

Answer 7

Awesome! and so for this one, would the limit = 2?

Answer 8

yes

Answer 9

you got a hang on the concept and now you can do these problems in a sec:)

Answer 10

absolute value of 2, is equivalent to 10*0 +2 ...

Answer 11

so this one would also be continous

Answer 12

ok:) so I got those questions down:P lol and would you be able to help me solve for vertical asymptotes?:)
Use graphs and tables to find the limit and identify any vertical asymptotes of

Answer 13

okay, solving for vertical asymptotes (saying, the x-values at which the function is undefined), using limits:

Answer 14

you have a left sides limit in the attachment, so that is the left part of the function.
It will approach negative infinity. Why? because if you plug a number that is less than 5 into the limit you get:

1 / (small negative decimal).

Answer 15

as you plug in numbers that approach 5 closer and closer (but their difference between 5 and the numbers you plug in, becomes a very small negative number like -0.0001

Answer 16

or even -0.00000001 and approaching 5 even closer, it gets to -0.000000001

Answer 17

ok, so would the horizontal asymptote be -infinity and the vertical asymptote be x=5?

Answer 18

so you will be getting
1 / -0.0001
1 / -0.0000001
1 / -0.000000001
which results in a larger magnitude negative infinity, like if you were to fill the table of values (numerical method, going from to the values that approach -5 closer and closer such as 4.99999999 and -4.9999...99999 and on then you would be getting

bigger negative infinity

Answer 19

yes, the vertical asymptote is at x=5

Answer 20

AWESOME lol im finally sorta getting precalc:p

Answer 21

and if you observe the other side, then you very small positive decimal nearly zero, and then the principle is the same

you get
1/ 0.0001
1/ 0.0000001
1/ 0.00000000001

and on, which results in larger positive values

Answer 22

or basically:
\[\LARGE \lim_{x \rightarrow 5^- } \frac{1}{x-5}=- \infty \]\[\LARGE \lim_{x \rightarrow 5 } \frac{1}{x-5}= \rm DNE\]\[\LARGE \lim_{x \rightarrow 5^+ } \frac{1}{x-5}=+ \infty \]

Answer 23

this is the case of the vertical asymptote at x=5.

Answer 24

sketch,