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Thread: limits

  1. #1
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    limits

    $\displaystyle lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

    $\displaystyle lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

    $\displaystyle lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

    $\displaystyle lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

    $\displaystyle lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

    can i verify the solution..thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    [quote=ashes;127090]$\displaystyle lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

    $\displaystyle lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

    $\displaystyle lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

    $\displaystyle lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

    $\displaystyle lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

    can i verify the solution..thanks...for the first one because of l'hopitals we have $\displaystyle \lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}$...applying L'hopitals rule again we get $\displaystyle \lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)}$ splitting we get $\displaystyle \lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2$...and finally differntiatting on the first one we et $\displaystyle \lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1$
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    [quote=Mathstud28;127109]
    Quote Originally Posted by ashes View Post
    $\displaystyle lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

    $\displaystyle lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

    $\displaystyle lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

    $\displaystyle lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

    $\displaystyle lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

    can i verify the solution..thanks...for the first one because of l'hopitals we have $\displaystyle \lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}$...applying L'hopitals rule again we get $\displaystyle \lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)}$ splitting we get $\displaystyle \lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2$...and finally differntiatting on the first one we et $\displaystyle \lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1$

    The second and third one are just more simple applications of L'hopitals if I have time tonight I will show them...here is the fourth one $\displaystyle \lim_{r \to 1}\frac{ar^{n}-a}{r-1}$....since it is $\displaystyle \frac{0}{0}$...we can apply L'hopital's rule to get $\displaystyle \lim_{r \to 1}\frac{anr^{n-1}}{1}$=an
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    And I will give you a hint for the last one

    Quote Originally Posted by ashes View Post
    $\displaystyle lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

    $\displaystyle lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

    $\displaystyle lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

    $\displaystyle lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

    $\displaystyle lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

    can i verify the solution..thanks
    if you combine them you have $\displaystyle \lim_{x \to 0}\frac{x-sin(x)}{xsin(x)}$....hmm....I wonder what you could use on that...*cough*L'hopital's*cough*
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