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Math Help - Calculating limit on a trig problem

  1. #1
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    Calculating limit on a trig problem

    Hi,
    Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

    \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}

    \lim_{x \rightarrow \infty} xsin \frac{1}{x}


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    Quote Originally Posted by Robb View Post
    Hi,
    Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

    \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}

    Note that \frac {\sin x^4}{\sin^2 x^2} = \frac {\sin x^4}{\sin^2 x^2} \cdot \frac {x^4}{x^4} = \frac {\sin x^4}{x^4} \cdot \left( \frac {x^2}{\sin x^2} \right)^2

    now take the limit. you should be able to take it from here
    \lim_{x \rightarrow \infty} xsin \frac{1}{x}

    Let t = \frac 1x. then we have \lim_{t \to 0} \frac {\sin t}t

    i suppose you can take it from here.
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  3. #3
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    Hello,
    Quote Originally Posted by Robb View Post
    Hi,
    Needing help on the following limits, its been a long time since I have done trig, and have forgotten allot of trig properties.. any help would be appreciated on workings.

    \lim_{x \rightarrow 0} \frac{sinx^4}{sin^2x^2}

    \lim_{x \rightarrow \infty} xsin \frac{1}{x}


    The key formula for these is \lim_{x \to 0} \frac{\sin(x)}{x}=1


    the first one can be written :
    \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2

    The second one can be solved by making the substitution t=1/x



    edit : lol too late ><
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    edit : lol too late ><
    well, you gave the key formula
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  5. #5
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    Ok, Well i think I have got it now for the first question...

    Using  \lim_{x \to 0} \frac{\sin(x)}{x}=1

    I then took
    \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot<br />
\frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot<br />
\left(\frac{x^2}{\sin(x^2)}\right)^2<br />

    Setting t = x^4
     lim_{t \to 0} \frac{\sin(t)}{t} \cdot<br />
lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2<br />
    is equal to

    <br />
\lim_{x \to 0} \ 1 \cdot
    \left(\frac{x^2}{\sin(x^2)}\right)^2<br />

    Setting t = x^2

    <br />
\lim_{t \to 0} \ 1 \cdot \left(\frac{t}{\sin(t)}\right)^2

    Taking the inverse:
     \lim_{t \to 0} \frac{1}{1} \cdot \left(\frac{\sin(t)}{t}\right)^2 =  \lim_{t \to 0} \ 1 \cdot (1)^2 = 1
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Robb View Post
    Ok, Well i think I have got it now for the first question...

    Using  \lim_{x \to 0} \frac{\sin(x)}{x}=1

    I then took
    \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot<br />
\frac{x^4}{\sin^2(x^2)}=\lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot<br />
\left(\frac{x^2}{\sin(x^2)}\right)^2<br />

    Setting t = x^4
     lim_{t \to 0} \frac{\sin(t)}{t} \cdot<br />
lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2<br />
    is equal to

    <br />
\lim_{x \to 0} \ 1 \cdot
    \left(\frac{x^2}{\sin(x^2)}\right)^2<br />

    Setting t = x^2

    <br />
\lim_{t \to 0} \ 1 \cdot \left(\frac{t}{\sin(t)}\right)^2

    Taking the inverse:
     \lim_{t \to 0} \frac{1}{1} \cdot \left(\frac{\sin(t)}{t}\right)^2 =  \lim_{t \to 0} \ 1 \cdot (1)^2 = 1
    yes. but you didn't have to drag it out like that, just make two substitutions at once (if necessary, if you just said the limit was 1, i doubt your professor would mind).

    \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \left(\frac{x^2}{\sin(x^2)}\right)^2 = \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2

    Now let t = x^4 and r = x^2, then we have

    \lim_{x \to 0} \frac{\sin(x^4)}{x^4} \cdot \lim_{x \to 0} \left(\frac{x^2}{\sin(x^2)}\right)^2 = \lim_{t \to 0} \frac{\sin t}t \cdot \lim_{r \to 0} \left(\frac r{\sin r}\right)^2 = 1 \cdot 1^2 = 1


    (By the way, you can't just take inverses like that)
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