limit (x csc x) is? as x approaches zero
cscx = 1/sinx
\[\lim_{x\to~0}\frac{sin~x}{x}=1\] therefore, what do you think the limit would be of \[\lim_{x\to~0}\frac{x}{sin~x}\]
thank you. but the answer you gave is not one of the choices
we arent spose to give direct answers, only guidance
okay. thank you.
do i plug in the zero in both of the x?
no, you should "know" that sin(x)/x has a limit of 1; and that x csc(x) is just x/sin(x) and compare the results
there is something to do with a squeeze thrm which i never really can implement, or a Lhopital
lol proving what amistre said would require squeeze therorem and that's a pain lol
theorem* we had to do that on our first test
:)
squeeze theorem isn't hard, just almost unnecessary as most of what it proves in many cases that we use it in can be considered kind of like rules.
so sin x/sinx = 1 is a rule?
well anything over itself is 1
i mean x/sinx. sorry
rule to memorize \[\lim_{x\to~0}\frac{sin~x}{x}=1\]
That is more like a law lol.
(the anything over itself bit)
0/0 aint 1 ;)
the one exception lol
Our professors don't really treat zero as a number sometimes lol. Anything over 0 is undefined
do you guys know increasing/decreasing using the first derivatives of a equation and concavity using 2nd deriv.?
and a limit with 0 as the denominator (with the exception of 0 as the numerator as well, which would entail L'Hopital's rule) is DNE
yes
do you have a specific question in mind about those, such as drawing a graph?
so lets say in a func the 1st deriv is greater than 0 for all x but the 2nd deriv is less than 0 for all x? how would a graph of that look like? and can you explain the reasoning behind you ans.
that means that it is positive, increasing, but concave down so something like this|dw:1354141662047:dw| forgive my poor mouse skills
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