Question

Assume that lim x->c = [f(x)+g(x)] = 2, lim x->c [f(x)-g(x)] = 1and that lim x->c f(x) and lim x->c g(x) exsist. find those limits?

Answer 1

f(c) + g(c)=2...............i
f(c) - g(c)=-1.............ii

Answer 2

Now, solve eqn i and ii to get f(c) and g(c)

Answer 3

DOES IT HELP?

Answer 4

opps
f(c) - g(c)=1.............ii

Answer 5

ur a goddam retard

Answer 6

Your head is broken dumbass

Answer 7

u is stupid

Answer 8

???

Answer 9

got it?

Answer 10

what are all those comments up there? ^

Answer 11

it isn't making sense? how will I solve?

Answer 12

\[\lim_{x\to c}[f(x)+g(x)] = 2\\\lim_{x\to c}[f(x)-g(x)] = 1\]
\[\lim_{x\to c}[f(x)+g(x)] = 2\\\lim_{x\to c}f(x)+\lim_{x\to c}[g(x)] = 2\]
\[\lim_{x\to c}[f(x)-g(x)] = 1\\\lim_{x\to c}f(x)-\lim_{x\to c}[g(x)] = 1\]
\[\lim_{x\to c}f(x)+\lim_{x\to c}[g(x)] = 2\\\lim_{x\to c}f(x)-\lim_{x\to c}[g(x)] = 1\\2\lim_{x\to c}f(x)=3\\\lim_{x\to c}f(x)=\frac32\]
\[\lim_{x\to c}f(x)-\lim_{x\to c}[g(x)] = 1\\\lim_{x\to c}g(x)=\frac32-1=\frac12\]