Finding limits using theorems

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In summary, by applying theorems such as the continuity of functions and the limits of trigonometric functions, we can solve for the limit of \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}})), which simplifies to \lim_{x\rightarrow 1} \cos(arctan(1)). Using the properties of trigonometric functions, we can further simplify this to cos(\pi/4) = \frac{\sqrt{2}}{2}.
  • #1
Numnum
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Homework Statement



Use theorems to find the limit:

[tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))
[/tex]


Homework Equations



Theorems like
[tex] f(x)=c [/tex] is continuous
[tex] f(x)=x [/tex] is continuous
[tex] \lim_{x\rightarrow 0} \cos(x)=1 [/tex]
[tex] \lim_{x\rightarrow 0} \sin(x)=0 [/tex]
[tex] \lim_{x\rightarrow a} \sin(x)=sin(a) [/tex]
[tex] \lim_{x\rightarrow 0} \sin(x-a)=0 [/tex]

The Attempt at a Solution



I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?
 
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  • #2
Numnum said:

Homework Statement



Use theorems to find the limit:

[tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))
[/tex]

Homework Equations



Theorems like
[tex] f(x)=c [/tex] is continuous
[tex] f(x)=x [/tex] is continuous
[tex] \lim_{x\rightarrow 0} \cos(x)=1 [/tex]
[tex] \lim_{x\rightarrow 0} \sin(x)=0 [/tex]
[tex] \lim_{x\rightarrow a} \sin(x)=sin(a) [/tex]
[tex] \lim_{x\rightarrow 0} \sin(x-a)=0 [/tex]

The Attempt at a Solution



I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?

Good idea! Let u=x-1. You should also have a theorem about the limit of sin(u)/u as u->0.
 
Last edited:
  • #3
What is ##lim_{x → 0} \frac{sin(x)}{x}##?

How does it relate to ##lim_{x → 1} \frac{sin(x-1)}{x-1}##?
 
  • #4
Is this right?

So I have:


1. [tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))
[/tex]


2. [tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))
[/tex]


3. [tex]
\lim_{x\rightarrow 1} \cos(arctan(1))
[/tex]

because of the theorem: [tex]\lim_{x\rightarrow 0}({\frac{\sin(x)}{x}}))=1[/tex]

4. [tex]\lim_{x\rightarrow 1} cos({\frac{π}{4}})[/tex]

5. [tex]={\frac{1}{√2}}[/tex]
 
  • #5
Numnum said:
Is this right?

So I have:


1. [tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))
[/tex]


2. [tex]
\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))
[/tex]


3. [tex]
\lim_{x\rightarrow 1} \cos(arctan(1))
[/tex]

because of the theorem: [tex]\lim_{x\rightarrow 0}({\frac{\sin(x)}{x}}))=1[/tex]

4. [tex]\lim_{x\rightarrow 1} cos({\frac{π}{4}})[/tex]

5. [tex]={\frac{1}{√2}}[/tex]

Yes, and you are using your 'continuous function' theorems after you've worked out the sin(u)/u part, yes?
 
  • #6
Personally, I would do it the other way around. Since cosine is continuous, for all x, [itex]\lim_{x\to 1} cos(f(x))= cos(\lim_{x\to 1} f(x))[/itex].

That is, from [itex]\lim_{x\to 1} cos(actan(\frac{sin(x- 1)}{x})[/itex] we look at [itex]\lim_{x\to 1}arctan(\frac{sin(x-1)}{x})[/itex]. And since arctan is continuous for all x we look at [itex]\lim_{x\to 1}\frac{sin(x-1)}{x-1}[/itex]. As you say, that last limit is 1 so we have [itex]cos(arctan(1))= cos(\pi/4)= \frac{\sqrt{2}}{2}[/itex]
 

What is the definition of a limit?

The limit of a function is the value that a function approaches as the input approaches a certain value or point. It represents the behavior of the function as the input gets closer and closer to the specified value.

What is the difference between a left-hand and right-hand limit?

A left-hand limit is the value that a function approaches as the input approaches a certain value from the left side. Similarly, a right-hand limit is the value that a function approaches as the input approaches a certain value from the right side.

What is the Squeeze Theorem and how is it used to find limits?

The Squeeze Theorem states that if two functions have the same limit at a certain point, and a third function is squeezed between them, then the third function also has the same limit at that point. This theorem can be used to evaluate limits by finding two functions that are greater and less than the given function, and have the same limit at the desired point.

What is the Intermediate Value Theorem and how is it used to find limits?

The Intermediate Value Theorem states that if a function is continuous on a closed interval, and the function takes on two different values at the endpoints of the interval, then it must also take on every value in between those two values. This theorem can be used to find limits by narrowing down the possible values of the limit through a process of elimination.

What are the common methods for finding limits using theorems?

Some common methods for finding limits using theorems include the Squeeze Theorem, the Intermediate Value Theorem, and the Sandwich Theorem. Other methods include direct substitution, factoring and canceling, and using trigonometric identities to simplify the function before evaluating the limit.

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