Can someone help me with my Calculus BC class please? I need to estimate the limit by substituting smaller and smaller values of h. lim[h:0,((3+h)^3-27)/(h)]
Does it tell you a suggested value of \(h\) to start with?
The question just says: Estimate the limits in Problems 11-14 by substituting smaller and smaller values of h. For trigonometric functions, use radians. Give answers to one decimal place. lim as h approaches 0 (3+h)^3-27 over h
Well, I would start with \(h=1\), and then \(h=.5\), then \(.1\), and then \(.01\), and then \(.001\),... or something like that. Basically just plug those different values of \(h\) in, and see what you get. Let me know what integer you seem to be approaching
Well I found the answer to be 27 but I've absolutely no idea how. Ha ha.
Let's make a table (accurate to 3 decimal places).\[\begin{array}{|c|c|} \hline h&\frac{(3+h)^3-27}{h}\\ \hline 1 & 37\\ \hline .5&31.75\\ \hline .1&27.91\\ \hline.05&27.453\\ \hline.01&27.090\\ \hline.005&27.045\\ \hline.001&27.009\\ \hline \end{array}\]
As you can see, as \(h\) is approaching 0, the formula is approaching \(27\). So you would guess that the limit is approaching is 27.
You don't have to do any factoring? At all? :O i.e., FOIL? and how exactly did you get this table?
You don't have to do any kind of factoring at all. Just plug in values and get numbers. As for the table, I calculated the values using wolfram alpha. Unless you're asking how I made the actual table. To get that, just copy/paste this into the equation editor. ``` \begin{array}{|c|c|} \hline h&\frac{(3+h)^3-27}{h}\\ \hline 1 & 37\\ \hline .5&31.75\\ \hline .1&27.91\\ \hline .05&27.453\\ \hline .01&27.090\\ \hline .005&27.045\\ \hline .001&27.009\\ \hline \end{array} ```
The table was made with \(\LaTeX\), and we even have a group dedicated to practicing it if you want to learn more. http://openstudy.com/study#/groups/LaTeX%20Practicing!%20%3A%29
I think I've got it. Thanks for the help dude :D
You're welcome.
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