Recent content by sean/mac

One mole of supercooled liquid tin (Sn) is adiabatically contained at 495K. Given: the melting temperature of Sn, Tm, Sn = 505 [K] the enthalpy of fusion of Sn, hfus, Sn = 7070 [J/mol] the heat capacity of liquid Sn, Cp, Sn(l) = 34.7 – 9.2 x10-3 T [J / (mol K)] the heat capacity of solid...

yeh after you do algebraic manipulation to get it in the indeterminate form of 2-2cos(x)-x^2 ----------------- (x^2)-(x^2)cos(x) you have to apply l'hospital's rule 4 times to get a non indeterminate form which is -1/6 the only problem is when i substitute in say 0.00001 or -0.00001...

[f(b) - f(a)]/[b-a]=f'(c) for b=x and a=0 gives [1/(1+x) - 1]/x=-1/(1+c)^2 which when i do the algebraic manipulation gives 1+x=(1+c)^2 i don't know how to make a relation between 1-x and 1/(1+x)

Let f(x)=1/(1+x) Use the Mean Value Theorom (for the derivative of a function) to prove that f(x)>=1-x for x>=0 also Mean Value Theorom states: [f(b)-f(a)]/ [b-a]= f'(c) where c is an element of [a,b]

lim (2/x^2)-(1/(1-cos(x))) x-->0 i have tried to use l'hopital's rule, but i keep on getting -1/6 however from graphing the function on my graphics calculator i know that it is equal to zero any help is appreciated

Give an example of a function f which is discontinuous at 0 yet abs(f ) is continuous at 0 i have tried for an hour or so trying to think of one, even hints would be helpful