(1/4 + 1/X)/(4+x) WITH LIMITAPPROACHING X TO -4
i cant use a calc
rewrite (1/4+1/x) so you can divide both sides by (4+x) and get rid of the indeterminant denom
4/4x + x/4x?
sure, whatever
just tell me what you get and I'll tell you if it's right
i already no the anwre but i dont know how to do it
\(\bf \Large lim_{x\rightarrow -4} \cfrac{ \frac{ \frac{1}{4}+1 }{x} }{ 4+x } \quad ? \)
dont btoher helpingme i ur trolling me
no its (1/x)
I'm not trolling you
then just get out
(1/4 + 1/X)/(4+x) = (16(4+4x)/(4+4x)) does that help?
the anwer isnt 16
ohh shoot... I see so.... \(\bf\Large lim_{x\rightarrow -4} \cfrac{ \frac{1}{4}+\frac{1}{x} } { 4+x } \) right
yes sir
http://www.wolframalpha.com/input/?i=%281%2F4+%2B+1%2FX%29%2F%284%2Bx%29++as+x+approaches+-4
ur toxic inky
@jdoe0001 can u help me plz?
Dude, I got a 5 on my AP BC calc test. also, disagreeing with a CAS is not very smart
the teacher put in -1/16 im in calc ab
i got -16
gimme a sec, I needa write down the problem if I want to make anything useful out of it.
\(\bf lim_{x\rightarrow -4} \cfrac{ \frac{1}{4}+\frac{1}{x} } { 4+x }\\ -------------\\ \cfrac{ \frac{1}{4}+\frac{1}{x} } { 4+x } \implies \cfrac{ \frac{x+4}{4x} }{4+x} \implies \cfrac{x+4}{4x} \times \cfrac{1}{4+x}\\ \implies \cfrac{\cancel{4+x}}{4x} \times \cfrac{1}{\cancel{4+x}}\)
yes but i need to know how they got the answer withough a calc?
yeah yeah I'll give the medal to that dude
lol
@morgan34 sign up and click show steps sigh
I'm too lazy to work out the problem since I don't have any paper with me, so you may as well get help from the other dude.
how come u didnt put 4 + x/1
cause you can't algebra
\(\bf\Large { \cfrac{ \frac{x+4}{4x} }{4+x} \implies \cfrac{ \frac{x+4}{4x} }{\frac{4+x}{1}}}\)
o yeah orgot ty
bless
dude, wolfram alpha works better - just get the 14 day pro trial and type in a problem and click show steps
cause then you can do ur maths homework in 3 minutes at 4AM
just don't make a habit out of it or you'll get a 1 on the AP test lol
mine expired
make a new account and do the same thing - free users can still get 3 free solutions a day
ok thanks
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