L'Hopital's Rule/Limit troubles

  • Thread starter Thread starter tangents
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the application of L'Hôpital's Rule in solving limits that yield indeterminate forms. The user struggles with two specific limits: lim x->0+ (Ln[sin[x]])/(Ln[sin[2x]]) and Lim x->0 (xcos[x]+e^-x)/x^2. For the first limit, the user incorrectly derives Cot[x]/2cot[2x] and fails to simplify it correctly, while for the second limit, the user miscalculates the limit after taking derivatives twice. The correct approach involves recognizing the behavior of the functions as x approaches 0 and simplifying the expressions accurately.

PREREQUISITES
  • Understanding of L'Hôpital's Rule for evaluating limits.
  • Knowledge of derivatives and their application in calculus.
  • Familiarity with trigonometric functions and their limits.
  • Basic logarithmic properties and their behavior near zero.
NEXT STEPS
  • Study the application of L'Hôpital's Rule with examples involving logarithmic functions.
  • Learn how to simplify trigonometric limits using identities.
  • Explore the behavior of functions near zero, particularly for sin(x) and cos(x).
  • Practice solving limits that result in indeterminate forms using higher-order derivatives.
USEFUL FOR

Students studying calculus, particularly those focusing on limits and derivatives, as well as educators looking for examples of common pitfalls in applying L'Hôpital's Rule.

tangents
Messages
21
Reaction score
0
Hello,
Well my math class finally started integrals after finishing H's Rule. But while I was doing my HW, I wouldn't get the correct answer but was sure I was following the rule. The rule is simple saying that if taking the limit gives indeteminate form ,then find dervative and then take limit. So anyway here are the two questions that I can't get
1) lim x->0+ (Ln[sin[x]])/(Ln[sin[2x]])
I took dervative and got Cot[x]/2cot[2x]. but since sin 0=0 i continued to get un defined so i took dervative again but got sin in the answer so i knew i was having some problem. The answer should be one but i can't seem to get it

2) Lim x->0 (xcos[x]+e^-x)/x^2
i did the dervative twice until the denominater was a constant and got the answer 0 but that isn't in domain. the correct answer is infinite but I again don't know how the textbook got that

Any ideas why I can't get the right answer =/
 
Physics news on Phys.org
As for 2), check the limit of your original numerator as x goes to 0.

As for 1), understand why we have:
[tex]\frac{\cot(x)}{2\cot(2x)}=\frac{\sin(2x)}{2\sin(x)}\frac{\cos{x}}{\cos(2x)}=\frac{\cos^{2}(x)}{\cos(2x)}[/tex]
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K