Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Can someone help me with this.. ? it's attached

OpenStudy (anonymous):

hartnn (hartnn):

\[\lim_{x \rightarrow 2}f(x)+g(x)=f(2)+g(2)\] can u find f(2) and g(2) ?

OpenStudy (anonymous):

is that what I do for the first one?

hartnn (hartnn):

yup, find f(2) and g(2) then add.

OpenStudy (anonymous):

is f(2)=1?

hartnn (hartnn):

nopes, see the graph, when x=2, y=f(x)= 2 ok?

OpenStudy (anonymous):

so, its the unshaded dot?

hartnn (hartnn):

oh, wait, i think the dot is shaded to make the value of f(2) as 1. and not 2 so f(2)=1 as u correctly said before.

OpenStudy (anonymous):

oh kay.. so first one is 3?

hartnn (hartnn):

but g(2) is 0 isn't it? so, 1+0=1 is the value of limit, ok?

OpenStudy (anonymous):

huh?

hartnn (hartnn):

didn't u get g(2) = 0 ??

OpenStudy (anonymous):

ohh yeah.. because theres no point

hartnn (hartnn):

thats because when x=2,g(x) is on x-axis, so g(2)=0

OpenStudy (anonymous):

for the second one is f(x)= 1?

OpenStudy (anonymous):

and g(x)=1?

hartnn (hartnn):

yup,both correct,so whats the value of limit?

OpenStudy (anonymous):

2?

hartnn (hartnn):

yup,thats correct,try 3rd.

OpenStudy (anonymous):

and the next one is 0 overall right?

OpenStudy (anonymous):

f(x)=0 and because its multiplying the total limit is 0

hartnn (hartnn):

right,because f(0)=0

hartnn (hartnn):

i think u are getting this,so just write what answers u get,i'll verify.

OpenStudy (anonymous):

d is -1/0=0

hartnn (hartnn):

anything/0 = infinity so d is -infinity

OpenStudy (anonymous):

oh..

hartnn (hartnn):

e is simpler,what did u get for e ?

OpenStudy (anonymous):

e=8?

hartnn (hartnn):

yup :)

OpenStudy (anonymous):

f=2?

hartnn (hartnn):

absolutely, good work :)

OpenStudy (anonymous):

yay!!! thanks!

hartnn (hartnn):

welcome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!