1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding limits using theorems

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Use theorems to find the limit:

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


    2. Relevant 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]

    3. 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?
     
  2. jcsd
  3. May 27, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good idea! Let u=x-1. You should also have a theorem about the limit of sin(u)/u as u->0.
     
    Last edited: May 27, 2013
  4. May 27, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

    What is ##lim_{x → 0} \frac{sin(x)}{x}##?

    How does it relate to ##lim_{x → 1} \frac{sin(x-1)}{x-1}##?
     
  5. May 27, 2013 #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]
     
  6. May 27, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, and you are using your 'continuous function' theorems after you've worked out the sin(u)/u part, yes?
     
  7. May 27, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding limits using theorems
  1. Limit using theorems (Replies: 2)

Loading...