Question

Limit x->7+ to (x-8)/(x-7)?

Answer 1

First, let's set up a table. To create the table, simply plug in values of x and record the corresponding values of y. For instance, when x = 6, then y = (6-8)/(6-7) = -2/(-1) = 2, so one line is 6 2


So do this for values near 7 to get


x y
6.00000 2.00000
6.10000 2.11111
6.20000 2.25000
6.30000 2.42857
6.40000 2.66667
6.50000 3.00000
6.60000 3.50000
6.70000 4.33333
6.80000 6.00000
6.90000 11.00000
7.00000 undefined
7.10000 -9.00000
7.20000 -4.00000
7.30000 -2.33333
7.40000 -1.50000
7.50000 -1.00000
7.60000 -0.66667
7.70000 -0.42857
7.80000 -0.25000
7.90000 -0.11111
8.00000 0.00000


Notice how as x decreases towards x=7, the y values are becoming larger and larger in the negative direction. If you keep going, these values will approach negative infinity.


So \[\large \lim_{x\to7^{+}}\left(\frac{x-8}{x-7}\right)=-\infty\]

Answer 2

Ohhhhhhhh! What if 7 is coming from the left? As x approaches 7-

Answer 3

then we look at the table for x values that are slowly increasing to x=7

these values point to y values that are increasing


so \[\LARGE \lim_{x\to7^{-}}\left(\frac{x-8}{x-7}\right)=\infty\]

Answer 4

Omg you taught me so much right now.. and since when x comes from the left is infinity and as it comes from the right its negative infinity there isnt a limit as x just apporaches 7... :D

Answer 5

exactly, the limit

\[\large \lim_{x\to7^{-}}\left(\frac{x-8}{x-7}\right)\]

doesn't exist because

\[\LARGE \lim_{x\to7^{-}}\left(\frac{x-8}{x-7}\right)\neq \lim_{x\to7^{+}}\left(\frac{x-8}{x-7}\right)\]

Answer 6

Thank you so much:)

Answer 7

\[\lim_{x \rightarrow 0}(3/x^2+3x)-(1/x)\]?