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Math Help - Simplifying limits

  1. #1
    Junior Member Misa-Campo's Avatar
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    Simplifying limits

    In general how would i simplify these expressions showing enough proof? and is there a website where i see all the rules for limits?


    1. Limit Sin ax/Sin bx
    x -->0

    2. Limit Tan ax/bx
    x -->0

    3. Limit Cos x/x
    x -->0
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    In general how would i simplify these expressions showing enough proof? and is there a website where i see all the rules for limits?


    1. Limit Sin ax/Sin bx
    x -->0

    2. Limit Tan ax/bx
    x -->0

    3. Limit Cos x/x
    x -->0

    Here how about this,


    Let f(x) and g(x) be two functions such that

    f(x)\sim{g(x)}\quad\text{As }x\to{c}

    Or in other words

    \lim_{x\to{c}}\frac{f(x)}{g(x)}=1


    Then you can replace f(x) with g(x) in any non-related limit that goes to c.

    Here I will show why

    Say we are still talking about f(x)\sim{g(x)}


    But we need to compute the limit \lim_{x\to{c}}\frac{f(x)}{h(x)}

    And if we could replace f(x) with g(x) this would be much simpler.

    So here is how we can see we can

    Since

    \lim_{x\to{c}}\frac{f(x)}{g(x)}=1

    From our laws of limits we see that

    \frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}g(x)}=1

    Assuming both limits exist, we then solve for the limit in question

    \lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x).


    Now you may be wondering how this helps, but once again consider that our limit we wish to compute may be rewritten as

    \frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}h(x)}

    Now we see by the equation above that we may rewrite this as

    \frac{\lim_{x\to{c}}g(x)}{\lim_{x\to{c}}h(x)}=\lim  _{x\to{c}}\frac{g(x)}{h(x)}

    Which is what we desired.
    --------------------------------------------------------------------------
    So now for the first example what you may do is this

    Since

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

    By the substitution ax=\varphi

    and similarly for \sin(bx)

    We may say that

    \sin(ax)\sim{ax}

    and

    \sin(bx)\sim{bx}


    So we may rewrite our limit as follows

    \lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}\sim\lim_{x  \to{0}}\frac{ax}{bx}=\frac{a}{b}
    -----------------------------------------------------------------------

    Similarly by a substitution of ax=\varphi you can show that

    \tan(ax)\sim{ax}\quad\text{As }x\to{0}

    Or just consider that \tan(ax)=\frac{\sin(ax)}{\cos(ax)}

    and as x\to{0}\Rightarrow\cos(ax)\to{1}


    So we can see that

    \lim_{x\to{0}}\frac{\tan(ax)}{bx}\sim\lim_{x\to{0}  }\frac{ax}{bx}=\frac{a}{b}
    --------------------------------------------------------------------------

    The last limit does not exist, to make it a little more clear we can see that

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

    since \cos0)=1

    So \cos(x)\sim{1}\quad\text{As }x\to{0}

    So we may rewrite this as

    \lim_{x\to{0}}\frac{\cos(x)}{x}\sim\lim_{x\to{0}}\  frac{1}{x}

    In which case it should be apparent the limit does not exist since

    \lim_{x\to{0^-}}\frac{1}{x}\ne\lim_{x\to{0^+}}\frac{1}{x}
    --------------------------------------------------------------------------

    Note the first two limits could have also been done as follows

    \lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}=\lim_{x\to  {0}}\frac{ax}{bx}\cdot\frac{\sin(ax)}{ax}\cdot\fra  c{bx}{\sin(bx)}

    Now rewriting this limit as the products of its individual components we see that we have

    \frac{a}{b}\cdot\lim_{x\to{0}}\frac{\sin(ax)}{ax}\  cdot\lim_{x\to{0}}\frac{bx}{\sin(bx)}

    The last two limits can be found by making the sbustitutions \varphi=ax and \psi=bx respectively.
    -------------------------------------------------------------------------

    and also we could rewrite

    \lim_{x\to{0}}\frac{\tan(ax)}{bx}=\lim_{x\to{0}}=\  lim_{x\to{0}}\frac{\sin(ax)}{\cos(ax)bx}

    Now rewriting it as

    \lim_{x\to{0}}\frac{\sin(ax)}{bx}\cdot\lim_{x\to{c  }}\frac{1}{\cos(ax)}=\lim_{x\to{0}}\frac{\sin(ax)}  {bx}=\frac{a}{b}\lim_{x\to{0}}\frac{\sin(ax)}{ax}

    where the aforementioned substitution should be made.
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  3. #3
    Junior Member Misa-Campo's Avatar
    Joined
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    Australia, Sydney
    Posts
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    Thanks

    Thanks alot man for your time and support, i understand clearly now
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