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Math Help - help with differentiation and limits

  1. #1
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    help with differentiation and limits

    Differentiate y = logx3 (base is x) ie. y^3 = x


    Evaluate lim xsin1/x using u=1/x
    x=>


    lim (√x - √5)/(x-5)
    x=>5



    thankyou so much to anyone who can help me with these questions !
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) y=\log_x3=\frac{\ln 3}{\ln x}

    and use the formula \left(\frac{a}{y}\right)'=-\frac{ay'}{y^2}

    2) \lim_{x\to\infty}x\sin\frac{1}{x}=\lim_{x\to\infty  }\frac{\sin\frac{1}{x}}{\frac{1}{x}}

    Let u=\frac{1}{x}. Then u\to 0 as x\to\infty.

    Then the limit is \lim_{u\to 0}\frac{\sin u}{u}=1

    3) Factorize the denominator: x-5=(\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})
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  3. #3
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    Differentiate y = logx3 (base is x) ie. y^3 = x
    Well, you're going to run into problems because y=\log[x]{3} does not equal y^{3}=x. Remember, that the logarithim with base "a" is asking you to find the power to raise it to get "b". In this case: x^{y}=3. Do you see what you do from there?

    Evaluate lim xsin1/x using u=1/x
    x=>∞
    If u=\frac{1}{x} and x=\frac{1}{u};

    Then we can rewrite this problem as \frac{sin(u)}{u}. Now, we need to look at the equation u=\frac{1}{x}. The limit as x approaches infinity of this equation is zero yes? Then we can say that the limit as x approaches infinity of xsin(\frac{1}{x}) is the same as the limit as u approaches 0 of \frac{sin(u)}{u}.

    Now this may not be any help if you haven't learned the squeeze theorem of haven't seen a limit of sin and x written as such.

    lim (√x - √5)/(x-5)
    x=>5
    Oh I hated these when I first started. Because they look so dastardly, but when the instructor did it - it all seemed so easy. And it is as this can be done in a few steps:

    \frac{\sqrt{x}-\sqrt{5}}{x-5}\Rightarrow
    \frac{\sqrt{x}-\sqrt{5}}{x-5}\\frac{\sqrt{x}+\sqrt{5}}{\sqrt{x}+\sqrt{5}}\Rig  htarrow

    Can you take it from here?
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