How to Solve a Limit Calculation Problem with Multiple Choice Options?

  • Thread starter Thread starter nick850
  • Start date Start date
  • Tags Tags
    Limit Test
Click For Summary
SUMMARY

The limit calculation problem presented involves evaluating the limit as t approaches 3 for the expression 1/(t^2-3t) - 2/(t^2-9). The correct answer is (c) -1/18. To solve this, one must manipulate the equation by finding a common denominator, which is t(t-3)(t+3), and simplifying the expression to eliminate the indeterminate form. The final simplified expression leads directly to the limit value.

PREREQUISITES
  • Understanding of limit calculations in calculus
  • Familiarity with algebraic manipulation of fractions
  • Knowledge of common denominators in rational expressions
  • Experience with evaluating limits involving indeterminate forms
NEXT STEPS
  • Study techniques for resolving indeterminate forms in limits
  • Learn about the properties of rational functions and their limits
  • Explore advanced limit evaluation methods, such as L'Hôpital's Rule
  • Practice problems involving limits with multiple choice options
USEFUL FOR

Students studying calculus, particularly those preparing for exams involving limit calculations, as well as educators looking for effective methods to teach limit evaluation techniques.

nick850
Messages
14
Reaction score
0

Homework Statement



This is a problem off a multiple choice practice test:
lim t->3 ( 1/(t^2-3t) - 2/(t^2-9) =

The solutions are:
(a) 0 (b) -1/9 (c) -1/18 (d)1 (e) 1/3

The correct answer is (c).

Can someone explain to me how to solve it? Any help would be very appreciated. You don't need to explain how to do limits. I just need to know how to manipulate the equation so that it's not indeterminate.

Homework Equations



NA

The Attempt at a Solution



I tried multiplying by the conjugate with no success.
 
Physics news on Phys.org
The first thing I would do is actually subtract the two fractions:
\frac{1}{t^2- 3t}- \frac{2}{t^2- 9}= \frac{1}{t(t- 3)}- \frac{2}{(t- 3)(t+ 3)}

Clearly the "common denominator" is t(t- 3)(t+ 3):
\frac{t+ 3}{t(t- 3)(t+ 3)}- \frac{2t}{t(t- 3)(t+ 3)}= \frac{t+ 3- 2t}{t(t- 3)(t+ 3)}= \frac{-t+ 3}{t(t- 3)(t+ 3)}= -\frac{t- 3}{t(t- 3)(t+ 3)}[/itex]
 
Thank you!
 

Similar threads

Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
35
Views
5K
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
17
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K