anonymous
  • anonymous
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)]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KingGeorge
  • KingGeorge
Does it tell you a suggested value of \(h\) to start with?
anonymous
  • anonymous
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
KingGeorge
  • KingGeorge
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Well I found the answer to be 27 but I've absolutely no idea how. Ha ha.
KingGeorge
  • KingGeorge
Let's make a table (accurate to 3 decimal places).\[\begin{array}{|c|c|} \hline h&\frac{(3+h)^3-27}{h}\\ \hline 1 & 37\\ \hline .5&31.75\\ \hline .1&27.91\\ \hline.05&27.453\\ \hline.01&27.090\\ \hline.005&27.045\\ \hline.001&27.009\\ \hline \end{array}\]
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
You don't have to do any factoring? At all? :O i.e., FOIL? and how exactly did you get this table?
KingGeorge
  • KingGeorge
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} ```
KingGeorge
  • KingGeorge
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
anonymous
  • anonymous
I think I've got it. Thanks for the help dude :D
KingGeorge
  • KingGeorge
You're welcome.

Looking for something else?

Not the answer you are looking for? Search for more explanations.