#1 using l'Hospital's Rule,
( t^(1/3) - 1)/(t^(1/2) - 1) -> (1/3 t^2/3)/(1/2 t^1/2)
= 2/3 t^-1/6
= 2/3 at t=1
If y=2^x,
y' = ln2 2^x
at x=0, that's just ln2
So, the line is
y-1 = ln2(x)
so, y = ln2 x + 1
So, the slope is ln2 = 0.693
See
http://www.wolframalpha.com/input/?i=plot+y%3D2^x%2C+y%3Dlog2+*+x+%2B+1
Or, you can evaluate the limit, but you wind up with the same value.
What steps got you to 0.597?
1. Evaluate the function at the given numbers (correct to six decimals places). Use the results to guess the value of the limit,or explain why it does not exist.
F(t)=( t^(1/3) - 1)/(t^(1/2) - 1) ;
t= 1.5,1.2,1.1,1.01,1.001;
The limit of F(t) as t approaches 1.
My work:
t=1.001, F(t)= 0.666611
t=1.01, F(t)=0.666114
t=1, F(t)= empty
t= 1.1, F(t)=0.661358
t=1.2,F(t)= 0.656488
t=1.5,F(t)=0.643905
The limit of F(t) as x approaches 1 is 0.66????
2. The slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) is the limit of (2^x-1)/x as x approaches 0. Estimate the slope to three decimal places.
My answer: 0.597?????
1 answer
1 answer