Results 1 to 6 of 6

Math Help - Trigonometric Limit Question

  1. #1
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86

    Trigonometric Limit Question

    Having problem with this one...tried out several different identities but

    lim x-> -pi/2

    1 + (tan x)^2/ (sec x)^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by SportfreundeKeaneKent View Post
    Having problem with this one...tried out several different identities but

    lim x-> -pi/2

    1 + (tan x)^2/ (sec x)^2
    Well, I'd transform it..

    \sec^2 x=\frac{1}{\cos^2 x} \implies \frac{1}{\sec^2 x}=\cos^2 x

    And \tan^2 x=\frac{\sin^2 x}{\cos^2 x}


    ------> \frac{1+\tan^2 x}{\sec^2 x}=\left(1+\frac{\sin^2 x}{\cos^2 x}\right) \cdot \cos^2 x

    Continue ?

    (I don't know if 1+ is in the fraction or not... But the trick would be the same)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    From
    Seoul, South Korea
    Posts
    128
    observe that [1+(tanx)^2]/(secx)^2=1/(secx)^2+(tanx)^2/(secx)^2

    this is equal to 1/(secx)^2 + [(sinx)^2/(cosx)^2]/[1/(cosx)^2]. (this is because (tanx)^2=(sinx)^2/(cosx)^2 and (secx)^2=1/(cosx)^2)

    here dividing by 1/(cosx)^2 is same as multiplying by (cosx)^2, which cancels out the denominator of first term.

    Hence we have that your original function is equal to [1/(secx)^2]+(sinx)^2. taking the limit of that as it approaches -pi/2, we can literally "plug in" -pi/2 into this function to see that 1/(secx)^2 = 1/2 at x=-pi/2 and (sinx)^2 = 1/2 at x=-pi/2. thus, adding it up we get that limit approachs 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by squarerootof2 View Post
    observe that [1+(tanx)^2]/(secx)^2=1/(secx)^2+(tanx)^2/(secx)^2

    this is equal to 1/(secx)^2 + [(sinx)^2/(cosx)^2]/[1/(cosx)^2]. (this is because (tanx)^2=(sinx)^2/(cosx)^2 and (secx)^2=1/(cosx)^2)

    here dividing by 1/(cosx)^2 is same as multiplying by (cosx)^2, which cancels out the denominator of first term.

    Hence we have that your original function is equal to [1/(secx)^2]+(sinx)^2. taking the limit of that as it approaches -pi/2, we can literally "plug in" -pi/2 into this function to see that 1/(secx)^2 = 1/2 at x=-pi/2 and (sinx)^2 = 1/2 at x=-pi/2. thus, adding it up we get that limit approachs 1.
    Or you can just remember that we get the famous identity :

    \cos^2 x+\sin^2 x=1

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by squarerootof2 View Post
    [snip]
    we can literally "plug in" -pi/2 into this function to see that 1/(secx)^2 = 1/2 at x=-pi/2 and (sinx)^2 = 1/2 at x=-pi/2. thus, adding it up we get that limit approachs 1.
    While it's true that the sum of these two limits is equal to 1, you might want to reconsider your stated values for each of the individual limits .......
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,868
    Thanks
    747
    Hello, SportfreundeKeaneKent!

    \lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x}
    How about this identity: . 1 + \tan^2\!x \:=\:\sec^2\!x

    Then: . \lim_{x\to\text{-}\frac{\pi}{2}}\frac{1 + \tan^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} \frac{\sec^2\!x}{\sec^2\!x} \;=\;\lim_{x\to\text{-}\frac{\pi}{2}} (1) \;=\;1

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. a trigonometric limit question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 11th 2010, 06:24 AM
  2. Trigonometric Limit Question
    Posted in the Calculus Forum
    Replies: 9
    Last Post: January 15th 2009, 12:58 AM
  3. Trigonometric limit...
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 15th 2008, 07:09 PM
  4. trigonometric limit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 30th 2006, 03:37 AM
  5. trigonometric limit
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 5th 2006, 11:30 PM

Search Tags


/mathhelpforum @mathhelpforum