Question

Find the limit as x->-1 of (x^2+1)/(x+1). i've forgotten how to figure out problems when denominator = 0

Answer 1

l' hospital rule,
differentiate the numerator n denominator

Answer 2

do that till u get a determinate form

Answer 3

never learned a hospital rule

Answer 4

can someone do steps pls?

Answer 5

ok you differentiate the numerator and the denominator and then substitute the value of x

Answer 6

no you can't use l'hospital here

Answer 7

why :O

Answer 8

\[(-1)^2+1=1+1=2 \neq 0\]

Answer 9

lol sorry :P mathematical error :P

Answer 10

can someone please just do a quick list of steps

Answer 11

so i can figure out others like it

Answer 12

the function is undefined at x=-1
this is not a hole so the limit does not exist

Answer 13

yes it does, the answer is -2

Answer 14

no the limit does not exist

Answer 15

you typed your wrong if the answer is -2

Answer 16

i did

Answer 17

my fault, fml

Answer 18

lol

Answer 19

x^2-1****

Answer 20

\[\lim_{x \rightarrow -1}\frac{x^2-1}{x+1}=\lim_{x \rightarrow -1}\frac{(x-1)(x+1)}{x+1}\]

Answer 21

lol myininaya u r too quick :P

Answer 22

damn

Answer 23

now you also use l'hospital if you wanted to since we do have 0/0 here

Answer 24

but i mean you don't have to
you can just use some algebra

Answer 25

like i did above (x+1)/(x+1)=1
so you are left with (x-1) just plug in -1

Answer 26

yeah

Answer 27

ty

Answer 28

Limit is +Infinity from the right and -Infinity from the left. Plot attached.

Answer 29

yeah limit at that point doesn't exist as left hand limit is not equal to right hand limit. But he made an error with the question its x^2-1 not x^+1