What is the value of k for continuity at x = π/4?

Click For Summary
SUMMARY

The value of k for continuity of the function f(x) = [1 - tan(x)]/[1 - √2 sin(x)] at x = π/4 is determined by ensuring that the limit as x approaches π/4 equals k/2. The limit can be evaluated by checking both sides of the equation, specifically using the expression lim x → π/4 f(x). The correct evaluation leads to the conclusion that k must equal 1 for the function to be continuous at this point.

PREREQUISITES
  • Understanding of limits in calculus
  • Knowledge of trigonometric functions, specifically tan(x) and sin(x)
  • Familiarity with continuity concepts in mathematical analysis
  • Ability to manipulate algebraic expressions involving limits
NEXT STEPS
  • Learn about evaluating limits using L'Hôpital's Rule
  • Study the properties of continuity in piecewise functions
  • Explore the behavior of trigonometric functions near critical points
  • Investigate the concept of removable discontinuities in calculus
USEFUL FOR

Students studying calculus, particularly those focusing on limits and continuity, as well as educators seeking to clarify these concepts for their students.

RandomGuy1
Messages
19
Reaction score
0

Homework Statement



f(x) = [1 - tan(x)]/[1 - √2 sin(x)] for x ≠ π/4
= k/2 for x = π/4

Find the value of k if the function is continuous at x = π/4

The Attempt at a Solution



This means that lim x → π/4 f(x) = k/2

I put x = (π/4 + h) and then evaluated the limit as h tended to zero. Doesn't work. Get sin (2h) in the denominator. Can I get a hint?
 
Physics news on Phys.org
RandomGuy1 said:

Homework Statement



f(x) = [1 - tan(x)]/[1 - √2 sin(x)] for x ≠ π/4
= k/2 for x = π/4

Find the value of k if the function is continuous at x = π/4

The Attempt at a Solution



This means that lim x → π/4 f(x) = k/2

I put x = (π/4 + h) and then evaluated the limit as h tended to zero. Doesn't work. Get sin (2h) in the denominator. Can I get a hint?

You should check both sides of the limit using ##\frac{1 - tan(x)}{1 - \sqrt{2} sin(x)}##.

This will allow you to figure out what ##k## must be in order for the function to truly be continuous at ##x = \pi/4##.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K