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Math Help - Limit calculation without Lhopital's rule

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    Limit calculation without Lhopital's rule

    Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.
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    MHF Contributor chisigma's Avatar
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    Re: Limit calculation without Lhopital's rule

    Let's start from the limit...

    \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1} (1)

    First we construct the Taylor expansions...

    e^{x}-e= e\ (x-1)\ \{1 + \frac{x-1}{2}+...\} (2)

    \sin (x^{2}-1) = (x-1)\ \{ (x+1) - \frac{(x-1)^{2}\ (x+1)^{3}}{3!} +...\} (3)

    ... and then we devide (2) and (3) obtaining the result...

    \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1}= \frace{e}{2} (4)




    Marry Christmas from Serbia

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    Member anonimnystefy's Avatar
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    Re: Limit calculation without Lhopital's rule

    hi chisigma

    it's not Christmas here in Serbia!

    and what about the second limit?
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    Re: Limit calculation without Lhopital's rule

    Oh, and i forgot to tell you the second condition, without using Taylor expansion.
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    MHF Contributor Siron's Avatar
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    Re: Limit calculation without Lhopital's rule

    Why do you want to solve them without using l'Hopital's rule or taylor expansion?
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    MHF Contributor Amer's Avatar
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by noteiler1 View Post
    Oh, and i forgot to tell you the second condition, without using Taylor expansion.
    and without substitution ?
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    Re: Limit calculation without Lhopital's rule

    I need to solve them without Lhopital's rule and Taylor expansion, becouse one of my friends has these problems during first semester of studying maths, and they don't know these technics yet. I solved couple of similar problems using the fact,
    that if you have \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} then you can substitute f(x) with h(x) when
    \lim_{x \to a} h(x) = 0 and
    \lim_{x \to a} \frac{f(x)}{h(x)}=1 and then calculate
    \lim_{x \to a} \frac{h(x)}{g(x)}. I can find these functions for
    \sin(\frac{x^2}{\pi}) and \sin(x^2 -1) but this doesn't help if i don't know these functions for 2^{\sqrt{\sin(x)+1}}-2
    and e^x - e
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  8. #8
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by noteiler1 View Post
    Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.
    \displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}
    @Prove It
    Did you notice that the whole point of this thread is that the limits cannot be done with L'Hospital's Rule nor with series?
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by noteiler1 View Post
    I need to solve them without Lhopital's rule and Taylor expansion, becouse one of my friends has these problems during first semester of studying maths, and they don't know these technics yet. I solved couple of similar problems using the fact,
    that if you have \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} then you can substitute f(x) with h(x) when
    \lim_{x \to a} h(x) = 0 and
    \lim_{x \to a} \frac{f(x)}{h(x)}=1 and then calculate
    \lim_{x \to a} \frac{h(x)}{g(x)}. I can find these functions for
    \sin(\frac{x^2}{\pi}) and \sin(x^2 -1) but this doesn't help if i don't know these functions for 2^{\sqrt{\sin(x)+1}}-2
    and e^x - e
    If you need help finding the derivative of \displaystyle \begin{align*} 2^{\sqrt{\sin{x} + 1}} \end{align*}, remember that \displaystyle \begin{align*} \frac{d}{dx}\left(a^x\right) &= a^x\ln{a} \end{align*}
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by noteiler1 View Post
    I can find these functions for \sin(\frac{x^2}{\pi}) and \sin(x^2 -1) but this doesn't help if i don't know these functions for 2^{\sqrt{\sin(x)+1}}-2 and e^x - e
    Use the linear approximation of f(x) by means of its tangent line. For example, for f(x)=e^x-e the equation of the tangent line at (1,0) is y=e(x-1) . So,

    \lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by FernandoRevilla View Post
    Use the linear approximation of f(x) by means of its tangent line. For example, for f(x)=e^x-e the equation of the tangent line at (1,0) is y=e(x-1) . So,

    \lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}
    Yes, indeed \lim_{x\to 1}\frac{e^x-e}{e(x-1)}=1 and
    \lim_{x \to 1} \frac{\sin(x^2-1)}{x^2-1}=1 so we can substitute and calculate \lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}. How couldn't i see it!! Thanks alot. But what about second problem?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by noteiler1 View Post
    How couldn't i see it!! Thanks alot. But what about second problem?
    Same way, find the corresponding tangent lines. Show your work (if you want), and we can check it.
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    Re: Limit calculation without Lhopital's rule

    Quote Originally Posted by Plato View Post
    @Prove It
    Did you notice that the whole point of this thread is that the limits cannot be done with L'Hospital's Rule nor with series?
    I must have misread the title...
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  15. #15
    MHF Contributor chisigma's Avatar
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    Re: Limit calculation without Lhopital's rule

    An alternative approach [I do hope it is 'allowed'...] to the second integral uses the 'infinite product'...

    \sin z = z\ \prod_{n=1}^{\infty} (1-\frac{z^{2}}{\pi^{2}\ n^{2}}) (1)

    Setting in (1) z=x^{2}-1 and with a little computation we obtain...

    \frac{e^{x}-e}{\sin (x^{2}-1)}= e\ \frac{e^{x-1}-1}{x-1}}\ \frac{1}{(x+1)\ \prod_{n=1}^{\infty} (1-\frac{(x^{2}-1)^{2}}{\pi^{2}\ n^{2}})} (2)

    ... and from (2), taking into account the 'fundamental limit'...

    \lim_{x \rightarrow 0} \frac{e^{x}-1}{x}=1 (3)

    ... we obtain [finally!]...

    \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1)}= \frac{e}{2} (4)




    Marry Christmas from Serbia

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