Estimate the limit by substituting smaller and smaller values ofh. Give your answer to three decimal places.

lim as h -> 0 for 3^h - 1 / h

much thanks

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- Sep 6th 2007, 02:53 PMmer1988Limits...numerically
Estimate the limit by substituting smaller and smaller values of

*h*. Give your answer to three decimal places.

lim as h -> 0 for 3^h - 1 / h

much thanks - Sep 6th 2007, 02:58 PMThePerfectHacker
- Sep 6th 2007, 03:09 PMmer1988
its seems to be approaching 1.1

- Sep 6th 2007, 03:17 PMThePerfectHacker
If you want to be extra accurate it is 1.099

- Sep 6th 2007, 03:26 PMmer1988
yeah, cuz it wanted it to three decimal places...thank you...

i have one more questoin, i got the first part but i need help with the other one...

t(sec) 0 0.2 0.4 0.6 0.8 1.0

s(ft) 0 0.1 1.4 3.4 6.2 9.6

the first part is 5.7ft/s the avg. velocity, then the second part asks, estimate the velocity of the car at t=0.6

then i dont know where to go - Sep 6th 2007, 03:39 PMKrizalid
That's a completely different question, give it its own thread.

- Sep 6th 2007, 03:43 PMtopsquark
There are two ways to answer this depending on how much work you want to put into it. (I'd recommend the longer way.)

The long way: Plot your points using t as your x axis and s as your y axis. Sketch a smooth curve connecting the points. Now, at t = 0.6 s draw a line tangent to the curve. The slope of this line is your speed at t = 0.6 s.

The quick way: It may be that your teacher hasn't taught you about that yet. So what I would do is find the average speed of the car between t = 0.4 s and t = 0.6 s, then again do it between t = 0.6 s and t = 0.8 s. Hopefully these two will be close because I'm about to (statistically speaking) screw things up. Average the two average speeds that you got. Taking the average of two averages is technically forbidden unless the graph is a straight line, but you should get a rough value doing this. (Something you might wish to do is to compare this "average average" with the average speed between the times t = 0.4 s and t = 0.8 s.)

-Dan - Sep 6th 2007, 03:49 PMmer1988
thanks...that worked, i was just findin the average between .4 and .6 and using that, i didnt know that doing it between .4 and .8 works too..

thanks again