PLEASE can some one attempt this: Given: as x goes to c for all functions[ f(x)=2; g(x)=0, h(x)=3] evaluate the limit as x goes to c for: h(x)/(x-c)
Hey Sat: I cannot figure out what c is:
I assume you have to use rules
well first of all we know that h(x) = 3 yes?
a) 3 b) - 2⁄3 c) 3⁄2 d) 0 e) does not exist
yes
3/(x-c)
i mean that is given
yes given
ok and we want the limit as x -> c yes?
yes
now the denominator obviously goes to 0
and we know for f(x) it is 2 and for g(x) it is 0
i don't care what c is, it doesn't matter. the limit as x -> c of x - c = 0 for sure
please explain why?
oh because as x goes to c, well... x goes to c. therefore x - c (the difference) goes to zero. it is just that
ok, got you!
lets imagine c was 5 and i asked you to compute the limit as x -> 5 of x - 5. you would say it is zero. at least i hope you would
that makes sense, intuitive actually
good
haha! I would!
therefore the limit does not exist as the denom is 0
so the limit is the denominator is 0 we have established that fact
in order for the limit of the whole fraction to exist, that means the limit in the numerator must also be zero. if it is not, there is no limit
i can go further if i substitute and get 0/0
maybe factor and cancel, or something more fancy
but if the limit in the numerator is a non - zero number, then i have no hope
i cannot go further if i see 2/0 for example. no limit
it was understanding that x goes to c, would also mean that x-c has to be zero - that is what stumped me
and since we know the numerator is identically 3, you are trying to compute the limit as x -> c of 3/(x-c) which does not exist
great explanation! should get bonus medal
well now you are not stumped i hope
def not, you are absolutely awesome!
i am stumped as to why this question included an f and a g. are there other parts maybe?
no other parts, think just to confuse
yeah i see that. ok good luck, think you are doing fine
thank you, been a long time since I did calc;
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