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    Good afternoon folks,

    I've been working few some problems sheets and have come a cropper with a question. Any help would be massively appreciated.

    \lim_{x \to \frac{\pi}{2}}(\sec x-\tan x)

    My attempt at this question led to me separating tanx into \frac{\sin x}{\cos x} which gives us a common denominator of \frac{1}{\cos x}, so we have \frac{1 - \sin x}{\cos x}

    This, however, led me down a bit of a dark alley, without a paddle, so to speak. I tried substituting the Maclaurin expansion in, which I couldn't simplify or evaluate.

    Where should I go?
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    try L'hopital rule
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    (Original post by nigel_s)
    Good afternoon folks,

    I've been working few some problems sheets and have come a cropper with a question. Any help would be massively appreciated.

    \lim_{x \to \frac{\pi}{2}}(\sec x-\tan x)

    My attempt at this question led to me separating tanx into \frac{\sin x}{\cos x} which gives us a common denominator of \frac{1}{\cos x}, so we have \frac{1 - \sin x}{\cos x}

    This, however, led me down a bit of a dark alley, without a paddle, so to speak. I tried substituting the Maclaurin expansion in, which I couldn't simplify or evaluate.

    Where should I go?
    You can use l'Hôpital's rule on \frac{1 - \sin x}{\cos x}.
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    Aha! Of course! Many thanks.

    I do have another question, I wonder if anyone could help with this one:

    \lim_{x \to \infty} \sqrt{x}(\sqrt{x + 1} - \sqrt{x})

    I really can't see where to go with this one. I tried taking the root x into the brackets, and then multiplying by \frac{\sqrt{x^2 + x}+\sqrt{x}}{\sqrt{x^2 + x}+\sqrt{x}} to give \frac{x^2}{\sqrt{x^2 + x}+\sqrt{x}} but I still can't see how to solve and suspect I've taken another wrong turn.

    Again, any help will be massively appreciated
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    (Original post by nigel_s)
    Aha! Of course! Many thanks.

    I do have another question, I wonder if anyone could help with this one:

    \lim_{x \to \infty} \sqrt{x}(\sqrt{x + 1} - \sqrt{x})

    I really can't see where to go with this one. I tried taking the root x into the brackets, and then multiplying by \frac{\sqrt{x^2 + x}+\sqrt{x}}{\sqrt{x^2 + x}+\sqrt{x}} to give \frac{x^2}{\sqrt{x^2 + x}+\sqrt{x}} but I still can't see how to solve and suspect I've taken another wrong turn.

    Again, any help will be massively appreciated
    Why not multiply by \dfrac{\sqrt{x+1}+\sqrt{x}}{ \sqrt{x+1} +\sqrt{x}} instead? I don't know why you multiplied by what you did -- it looks like you half-expanded the bracket or something... but either way it was wrong :p:
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    You've taken it into the brackets incorrectly. You should get \sqrt{x^2+x}-x

    Try multiplying through by \sqrt{x+1}+\sqrt{x} (on the top and bottom)
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    Okay have multiplied by \sqrt{x+1} + \sqrt{x}

    Giving me:
    \frac{\sqrt{x}}{\sqrt{x+1} + \sqrt{x}}

    Can't see what to do next though?
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    (Original post by nigel_s)
    Okay have multiplied by \sqrt{x+1} + \sqrt{x}

    Giving me:
    \frac{\sqrt{x}}{\sqrt{x+1} + \sqrt{x}}

    Can't see what to do next though?
    Divide top and bottom by \sqrt{x}, and then take the limit as x \to \infty.
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    Excellent, many thanks. Have solved both these problems now.

    However, I have one final question (for tonight, at least).

    \lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}

    What I did was to subsitute \tan^2 x for \frac{1 - 2\tan x}{\tan 2x}

    Which gave:

    \lim_{x \to \pi} \frac{\tan 2x(1 + \cos x)}{1 - 2\tan x}

    Which I calculate to be \frac{0}{1}=0

    However, the correct answer is \frac{1}{2}

    What've I done wrong?
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    (Original post by nigel_s)
    Excellent, many thanks. Have solved both these problems now.

    However, I have one final question (for tonight, at least).

    \lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}
    Start by using the identity 1+\tan^2 x \equiv \sec^2 x and express the whole lot in terms of \cos x. Then do some simplification and it falls out.
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    (Original post by nuodai)
    Start by using the identity 1+\tan^2 x \equiv \sec^2 x and express the whole lot in terms of \cos x. Then do some simplification and it falls out.
    Done - great, thanks very much for all your help
 
 
 
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