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Limit Question

  1. Feb 28, 2016 #1
    I have a limit problem. This is the problem:

    [tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]

    The solution is
    [tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]
    [tex]= \lim_{x - 2 \to 0} \frac{\tan [(2 - \sqrt{2x}) × \frac{2 + \sqrt{2x}}{2 + \sqrt{2x}} ]}{x(x - 2)}[/tex]
    [tex]= \lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]

    Let ##p = x - 2##

    and ##2 + \sqrt{2x} = 2 + \sqrt{2⋅2} = 2 + \sqrt{4} = 2 + 2 = 4##

    Then

    [tex]\lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]
    [tex]= \lim_{p \to 0} \frac{\tan [-\frac{1}{2}p]}{2p}[/tex]
    [tex]= -\frac{1}{4}[/tex]

    Is this correct?

    [NOTE: moved to this forum by mentor hence no homework template]
     
    Last edited by a moderator: Feb 29, 2016
  2. jcsd
  3. Feb 28, 2016 #2

    Ssnow

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    The result seems correct, but there are intermediate steps not well justified ...
     
  4. Feb 28, 2016 #3

    Ssnow

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    you can directly write ##\frac{\tan{(2-\sqrt{2x})}}{x^2-2x}=\frac{\tan{(2-\sqrt{2x})}(2-\sqrt{2x})}{(2-\sqrt{2x})(x^2-2x)}##, now ##\lim_{x\rightarrow 2}\frac{\tan(2-\sqrt{2x})}{(2-\sqrt{2x})}=\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1##, with ##t=2-\sqrt{2x}## ... so what remains is ##\lim_{x\rightarrow 2}\frac{(2-\sqrt{2x})}{x^2-2x}## ...
     
  5. Feb 28, 2016 #4
    I don't understand. Is ##\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{2 - \sqrt{2x}} = 1##?
     
  6. Feb 29, 2016 #5

    Ssnow

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    yes
     
  7. Feb 29, 2016 #6
    @Ssnow ... the answer is -1/4. Check your calculations again.

    @gede... the answer you have received is correct. Instead of doing all this, why don't you directly use the L-Hospital's rule?
     
  8. Feb 29, 2016 #7
    How exactly this L'Hospital's rule like?

    Please show me the work of it.
     
  9. Feb 29, 2016 #8

    Ssnow

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    When in the limit you find indefinite forms as ##\frac{\infty}{\infty},\frac{0}{0}##, you can derive the numerator and the denominator repeating the limit ... for the precise statement see: https://en.wikipedia.org/wiki/L'Hôpital's_rule

    you can apply this to the limit ##\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2-2x}## obtaining the result ...

    PS. The method more immediate but you must be familiar with the concept of derivation.
     
  10. Feb 29, 2016 #9
    @Ssnow you mean differentiation, not derivation right?
     
  11. Feb 29, 2016 #10

    Ssnow

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  12. Feb 29, 2016 #11
    Isn't ##\lim \frac{\tan t}{t} = 1## only valid if ##t \to 0##?

    So, why does ##\lim_{t \to 2} \frac{\tan t}{t} = 1##?
     
  13. Mar 1, 2016 #12

    Ssnow

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    Hi, I didn't write ##\lim_{t\rightarrow 2}\frac{\tan{t}}{t}=1##, I wrote ##\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1##. In fact let us ##t=2-\sqrt{2x}## then when ##x\rightarrow 2## we have that in the new variable ## t\rightarrow 0##... (if you change variable in the limit also the interested point change but the limit remain the same ...)
     
  14. Mar 1, 2016 #13
  15. Mar 1, 2016 #14

    Ssnow

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    In example 6 and 7 pag 106, limits are calculated with substitution methods
     
  16. Mar 1, 2016 #15
    I'm still confuse. How do you get ##-\frac{1}{4}## ?
     
  17. Mar 1, 2016 #16

    Ssnow

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    Ok, I show you the complete resolution that doesn't involve differentiation, we start with

    [tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]

    we multiply and divide by the factor ## 2-\sqrt{2x}## so will be:

    [tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}\frac{2-\sqrt{2x}}{x^2 - 2x}[/tex]

    now the limit of a product is the product of limits.
    We examine the first limit ##\lim_{x\rightarrow 2}\frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}## doing the substitution ##t=2 - \sqrt{2x}##. In this case as ##x\rightarrow 2## we have ##t\rightarrow 0## and the limit is:

    ##\lim_{x\rightarrow 2}\frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}=\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1##

    So the first limit is ##1##. Now we shall examine the second ##\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2 - 2x}##. We start adding and subtracting ##x##, so

    ##\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2 - 2x}=\lim_{x\rightarrow 2}\frac{2-x+x-\sqrt{2x}}{x^2 - 2x}=##

    splitting the ratio, we have:

    ##=\lim_{x\rightarrow 2}\frac{2-x}{x^2 - 2x}+\frac{x-\sqrt{2x}}{x^2 - 2x}=##

    we simplify and rewrite ##x^2 - 2x=(x-\sqrt{2x})(x+\sqrt{2x})## (we used ##A^2-B^2=(A-B)(A+B)##), so

    ##=\lim_{x\rightarrow 2}\frac{2-x}{x(x-2)}+\frac{x-\sqrt{2x}}{(x-\sqrt{2x})(x+\sqrt{2x})}=\lim_{x\rightarrow 2}\frac{-1}{x}+\frac{1}{(x+\sqrt{2x})}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}##

    Now ## 1\cdot \left(-\frac{1}{4}\right)=-\frac{1}{4}## and the total limit is ##\frac{-1}{4}##.
     
  18. Mar 1, 2016 #17
    Can you show me from inside above calculus textbook about ##\lim_{x \to 0} \frac{\tan x}{x} = 1##?

    Is ##\lim_{x \to 0} \frac{\cos x}{x} = 1##?
     
  19. Mar 1, 2016 #18

    Ssnow

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    ok, in your book is proved that ##\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1##, for the limit above we observe that:

    ##\lim_{x\rightarrow 0}\frac{\tan{x}}{x}=\lim_{x\rightarrow 0}\frac{\sin{x}}{x\cos{x}}=\lim_{x\rightarrow 0}\frac{\sin{x}}{x}\lim_{x\rightarrow 0}\frac{1}{\cos{x}}=1\cdot 1=1##

    (the second is not true ##\lim_{x\rightarrow 0}\frac{\cos{x}}{x}## does not exists...)
     
  20. Mar 1, 2016 #19
    @gede e I suggest you go through the basics of limits before attempting the question again. Feel free to ask any doubts. You can PM me or Ssnow.
     
  21. Mar 1, 2016 #20

    Mark44

    Staff: Mentor

    Sounds like a good idea.
     
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