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Math Help - Limits using Trig Functions

  1. #1
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    Unhappy Limits using Trig Functions

    Can someone help me with this: Find the limit if the limit exists:

    lim x-tanx
    x->0 sinx



    lim 1 - sine^2 (1)^2
    x->-2- t (t)
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  2. #2
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    Quote Originally Posted by Roxanne View Post
    Can someone help me with this: Find the limit if the limit exists:

    lim x-tanx
    x->0 sinx

    Mr F says: {\color{red} \lim_{x \rightarrow 0} \frac{x - \tan x}{\sin x} = \lim_{x \rightarrow 0} \left( \frac{x}{\sin x} - \frac{1}{\cos x} \right) = \lim_{x \rightarrow 0} \frac{x}{\sin x} - \lim_{x \rightarrow 0} \frac{1}{\cos x}}

    lim 1 - sine^2 (1)^2 Mr F says: This is incomprehensible.
    x->-2- t (t)
    ..
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  3. #3
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    Limits using trig functions

    find limit if limit exist:

    lim 1/t^2 sin^2 (t/2)
    x->0
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  4. #4
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    Hello, Roxanne!

    If no one is answering, there is confusion in your statement . . .


    I must assume it says: . \lim_{x\to0}\frac{\sin^2\!\frac{x}{2}}{x^2}


    In the denominator, multiply by \frac{4}{4}

    . . \lim_{x\to0}\,\frac{\sin^2\!\frac{x}{2}}{\frac{4}{  4}\cdot x^2} \;=\;\lim_{x\to0}\:\frac{\sin^2\!\frac{x}{2}}{4\cd  ot\frac{x^2}{4}} \;=\;\lim_{x\to0}\:\frac{1}{4}\cdot\frac{\sin^2\!\  frac{x}{2}}{(\frac{x}{2})^2} . = \;\frac{1}{4}\cdot\lim_{x\to0} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \;=\;\frac{1}{4}\cdot1^2 \;=\;\frac{1}{4}

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  5. #5
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    Quote Originally Posted by Roxanne View Post
    find limit if limit exist:

    lim 1/t^2 sin^2 (t/2)
    x->0
    Do you mean \frac{1}{t^2} \, \sin^2 \frac{t}{2} ? If so note that:

    \frac{1}{t^2} \, \sin^2 \frac{t}{2} = \frac{\sin^2 \frac{t}{2}}{t^2} = \frac{1}{4} \, \left( \frac{\sin \frac{t}{2}}{\frac{t}{2}} \right) \, \left( \frac{\sin \frac{t}{2}}{\frac{t}{2}} \right).
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