#1.
f(3.000000) = 37.000000
f(3.900000) = 87.859000
f(3.990000) = 94.271599
f(3.999000) = 94.927016
f(3.999900) = 94.992700
f(3.999990) = 94.999270
f(3.999999) = 94.999927
Looks like lim = 95, eh?
do #2 likewise
for lim sin(x)/x use L'Hopital's Rule
lim sinx/x = lim cosx/1 = 1
---------------------
lim [1 + (1/x)]^x ...... it is understood that x approaches infinity
= lim e^ln( [1 + (1/x)]^x )
= e^ ( lim ln( [1 + (1/x)]^x ))
= e^ ( lim x * ln[1 + (1/x)] )
= e^ ( lim ln[1 + (1/x)] / (1/x) ) .. the limit is now 0/0 ...
..... by L'Hopitals Rule
= e^ ( lim {1/[1 + 1/x] *(-1/x^2)} / {-1/x^2} )
= e^ lim 1/[1+1/x]
= e^1
= e.
Work out the problem and explain your steps as you go please!
1. Find the limit by numerical evidence of limx -->4 x3+4x2-7x-5
2. Find the limit of limx-->-9 x2-81x+9.
3. Find the limit of each of the following:
a. limx-->0 x+4x-2x
b. limx-->0 x3+51 x2-17
c. limn-->infinity 1+1nn
d. limx-->0 sinxx
1 answer
3 answers