Solving the Limit of sin(cos(x))/sec(x) as x->0

  • Thread starter Miike012
  • Start date
  • Tags
    Limit
In summary, the problem involves finding the limit of sin(cos(x))/sec(x) as x approaches 0. The answer in the book is sin(1), but the conversation discusses why this answer is correct and why using the sin(x)/x rule may not apply in this case. Ultimately, it is determined that cos(x) approaches 1 as x approaches 0, making sin(1) the correct answer.
  • #1
Miike012
1,009
0
Problem:

Lim sin(cos(x))/sec(x)
x -> 0

The answer in the book is sin(1).. which is obvious... but why do I not arrive at the same answer doing is this way...

Knowing as x approaches zero, sin(x)/x approaches 1...

( Sin(cos(x))/cos(x) )(cos^2(x)) = (1)(1) = 1 = / = sin(1)...?
 
Physics news on Phys.org
  • #2
Miike012 said:
Knowing as x approaches zero, sin(x)/x approaches 1...

( Sin(cos(x))/cos(x) )(cos^2(x)) = (1)(1) = 1 = / = sin(1)...?

The bolded part is correct. However, what is cos(x) approaching as x approaches 0?
 
  • #3
cos(x) approaches 1..
 
  • #4
Ok, so since cos(x) goes to 1:

[tex]\lim_{x \to 0} \frac{\sin(\cos(x))}{\cos(x)}[/tex]

is equivalent to:

[tex]\lim_{u \to 1} \frac{\sin(u)}{u}[/tex]

We can't apply the sin(_)/_ rule here. The term inside of the sine approaches 1.
 
  • #5
Thank you.
 
  • #6
I wasn't aware that you can't use that rule if the function actually approaches a value.
 
  • #7
Miike012 said:
I wasn't aware that you can't use that rule if the function actually approaches a value.

[tex]\lim_{x \to 0}\frac{\sin(x)}{x} = 1[/tex]

x needs to approach 0. If it approaches any other value (or does not exist), you can't use this identity.
 
  • #8
that makes sense. Thank you.
 

1. What is the limit of the function sin(cos(x))/sec(x) as x approaches 0?

The limit of the function sin(cos(x))/sec(x) as x approaches 0 is equal to 1.

2. How do you solve the limit of sin(cos(x))/sec(x) as x approaches 0?

To solve this limit, you can use the trigonometric identity cos(x) = 1 - 2sin^2(x/2) and the property lim x->0 sin(x)/x = 1. This will allow you to rewrite the function as (1 - 2sin^2(x/2))/(1/cos(x)), which simplifies to (1 - 2sin^2(x/2))cos(x). Then, using the squeeze theorem, you can show that the limit is equal to 1.

3. Can you use L'Hopital's rule to solve this limit?

No, L'Hopital's rule cannot be used in this case because both the numerator and denominator approach 0 as x approaches 0. This makes it an indeterminate form and L'Hopital's rule only applies to forms such as 0/0 or infinity/infinity.

4. What does the limit of sin(cos(x))/sec(x) as x approaches 0 represent?

This limit represents the behavior of the function near the point x = 0. It tells us the value that the function approaches as x gets closer and closer to 0.

5. Can you use a graph to visualize the limit of sin(cos(x))/sec(x) as x approaches 0?

Yes, you can use a graph to visualize the behavior of the function near the point x = 0. The graph will show that as x approaches 0, the function approaches a horizontal asymptote at y = 1.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
346
  • Calculus and Beyond Homework Help
Replies
10
Views
942
  • Calculus and Beyond Homework Help
Replies
1
Views
493
  • Calculus and Beyond Homework Help
Replies
2
Views
917
  • Calculus and Beyond Homework Help
Replies
1
Views
288
  • Calculus and Beyond Homework Help
Replies
6
Views
710
  • Calculus and Beyond Homework Help
Replies
3
Views
563
  • Calculus and Beyond Homework Help
Replies
8
Views
920
  • Calculus and Beyond Homework Help
Replies
2
Views
865
  • Calculus and Beyond Homework Help
Replies
8
Views
236
Back
Top