Question

i have a question in one sided limits ? anyone to help . i can post the question

Answer 1

anyone pl

Answer 2

the first question

Answer 3

@Preetha

Answer 4

@hartnn

Answer 5

whats your doubt ?
you can just plug in x=2 in left and right side functions and if the value is same, the limit exist

Answer 6

is it that all about ?

Answer 7

arent we supposed to do like this ?

Answer 8

f is continuous if left side limit = right side limit

Answer 9

thank you ! so just plug in the value ? thats it ?

Answer 10

yes,
you just need to select the proper function

like for \(x \to 2^-\)
since x<2, you will select "x-4" and then plug in x=2
(you can directly substitute as you are not getting any indeterminate form)

for \(x \to 2^+\), you'll select x^2-6
and plug in x=2

Answer 11

we need to plug in x=2, as we are considering the continuity at x=2

apart from x=2, the function is continuous everywhere,
because x-4 and x^2-6 are continuous everywhere

Answer 12

Thank you so much @hartnn i got it..

Answer 13

welcome ^_^
you seem to be new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 14

yes i just logged it,,, Thank you for the help and many more questions waiting,,!!

Answer 15

bring it on! :)

Answer 16

question 20 pl

Answer 17

@hartnn

Answer 18

1/x is infinity at x=0+
and -infinity at x=0-

so its not continuous at x=0 only

Answer 19

Thank you !!

Answer 20

so apart from x=0, its continous at all values ?

Answer 21

in the question , they never asked about x=0 right ?

Answer 22

yes.
oh yeah, then f is continuous everywhere

Answer 23

What will be the proper answer to this ? when i get to write at exams

Answer 24

i mean the step by step procedure @hartnn

Answer 25

You just state that the domain is all real numbers except 0