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Math Help - l'hopitals rule checks and help

  1. #1
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    l'hopitals rule checks and help

    1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

    2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

    3: The size of an animalís pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
    be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

    Can someone please check my answers for 1 and 2 and give me some help for question 3?

    P.S sorry about the bad notation for the equation and limits
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

    2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

    3: The size of an animalís pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
    be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

    Can someone please check my answers for 1 and 2 and give me some help for question 3?

    P.S sorry about the bad notation for the equation and limits
    \lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}

    To simplify things a bit we see can see a few things

    Firstly you can semi-simplify things by knowing that as x\to{0} \sin(x)\sim{x}

    which gives \lim_{x\to{0}}\frac{\sin(\sin(x))}{x}=\lim_{x\to{0  }}\frac{\cos(\sin(x))\cdot{\cos(x)}}{1}=1

    Or more generally \lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}=1,if\text{ }u(0)=0

    so from that we see that as x\to{0} , \sin(u(x))\sim{u(x)}

    so we could just say as x\to{0} , \sin(\sin(x))\sim{\sin(x)}

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



    For the second one apply the squeeze theorem

    \forall{x}>0,\frac{\ln(x)}{x^2}\leq\frac{1}{x}

    and \forall{x}>0,\frac{1}{x^2}\leq\frac{\ln(x)}{x^2}

    so \frac{1}{x^2}<\frac{\ln(x)}{x^2}<\frac{1}{x}

    so maintaing equality we introduce the limits

    \lim_{x\to\infty}\frac{1}{x^2}\leq\lim_{x\to\infty  }\frac{\ln(x)}{x^2}\leq\lim_{x\to{0}}\frac{1}{x}

    or

    0\leq\lim_{x\to{0}}\frac{\ln(x)}{x^2}\leq{0}

    \therefore\lim_{x\to{0}}\frac{\ln(x)}{x^2}=0 by the squeeze theorem
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  3. #3
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    Quote Originally Posted by deragon999 View Post
    1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

    Mr F says: Correct. But no need for the Hospital. Just make the substitution w = sin x. Note that as x \rightarrow 0, ~ w \rightarrow 0. ~ \lim_{w \rightarrow 0} \frac{\sin w}{w} is a standard result.


    2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

    Mr F says: Correct.


    3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
    be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

    Can someone please check my answers for 1 and 2 and give me some help for question 3?

    P.S sorry about the bad notation for the equation and limits
    3. You have f(x) = \frac{160 x^{-0.4} + 90}{8 x^{-0.4} + 10} = \frac{160 + 90 x^{0.4}}{8 + 10 x^{0.4}} = 9 + \frac{44}{5 x^{0.4} + 4} and the domain is x > 0.

    Note that as x \rightarrow 0^+, x^{0.4} \rightarrow 0.

    Note that as x \rightarrow +\infty, x^{-0.4} = \frac{1}{x^{0.4}}\rightarrow 0.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

    2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

    3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
    be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

    Can someone please check my answers for 1 and 2 and give me some help for question 3?

    P.S sorry about the bad notation for the equation and limits
    For the last one you are asked to find firstly

    \lim_{x\to\infty}\frac{160x^{-.4}+90}{8x^{-.4}+10}=\lim_{x\to\infty}\frac{\frac{160}{x^{.4}}+  90}{\frac{8}{x^{.4}}+10}<br />

    Direct substitution yields

    \frac{0+90}{0+10}=9


    And secondly you are asked to find

    \lim_{x\to{0}}\frac{\frac{160}{x^{.4}}+90}{\frac{8  }{x^{.4}}+10}

    Multiplying by \frac{x^{.4}}{x^{.4}}

    we get

    \lim_{x\to{0}}\frac{160+90x^{.4}}{8+10x^{.4}}

    now direct substitution yields

    \frac{160+0}{8+0}=20


    Now to find if they are the smallest/largest

    show that they are the absolute max and min points on

    (0,\infty)
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    and \forall{x}>0,\frac{1}{x^2}\leq\frac{\ln(x)}{x^2}

    Hmmm....but x \leq e \Rightarrow \ln x \leq 1
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by deragon999 View Post
    1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

    2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

    3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
    be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

    Can someone please check my answers for 1 and 2 and give me some help for question 3?

    P.S sorry about the bad notation for the equation and limits
    #1: \text{let} \ u=sin(x) and u(0)=0. Therefore, the limit is written as \lim_{u\to{0}}\frac{sin(u)}{u}=\color{red}\boxed{1  }.

    #2 \lim_{x\to{\infty}}\frac{\ln(x)}{x^2}.

    We get the indeterminate case \frac{\infty}{\infty}.

    Apply L'Hopital's Rule:

    \lim_{x\to{\infty}}\frac{\ln(x)}{x^2}=\lim_{x\to{\  infty}}\frac{\frac{1}{x}}{2x}=\lim_{x\to{\infty}}\  frac{1}{2x^2}=\color{red}\boxed{0}

    #3 \lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}

    If we plug in zero, we get the indeterminate case \frac{\infty}{\infty}

    Simplifying the limit, we get:

    \lim_{x\to{0^+}}\frac{\frac{160}{x^{.4}}+\frac{90x  ^{.4}}{x^{.4}}}{\frac{8}{x^{.4}}+\frac{10x^{.4}}{x  ^{.4}}}=\lim_{x\to{0^+}}\frac{\frac{1}{x^{.4}}}{\f  rac{1}{x^{.4}}}\cdot\frac{160+90x^{.4}}{8+10x^{.4}  }.

    Assuming that \frac{1}{x^{.4}}\neq 0, we have:

    \lim_{x\to{0^+}}\frac{160+90x^{.4}}{8+10x^{.4}}.

    Evaluating the limit, we now have: \frac{160}{8}=\color{red}\boxed{20}

    Now evaluate this as x\to{\infty}.

    \lim_{x\to{\infty}}\frac{160+90x^{.4}}{8+10x^{.4}}  =\frac{90}{10}=\color{red}\boxed{9}.

    Hope this helped you out!
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Hmmm....but x \leq e \Rightarrow \ln x \leq 1
    Oh how embarrasing! My calculator apparently cant solve inequalities well!
    Well how about this?
    \forall{x}>10^{10},\frac{1}{x^2}<\frac{\ln(x)}{x^2  }

    Happy?
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    \lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}

    To simplify things a bit we see can see a few things

    Firstly you can semi-simplify things by knowing that as x\to{0} \sin(x)\sim{x}

    which gives \lim_{x\to{0}}\frac{\sin(\sin(x))}{x}=\lim_{x\to{0  }}\frac{\cos(\sin(x))\cdot{\cos(x)}}{1}=1

    Or more generally \lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}=1,if\text{ }u(0)=0

    so from that we see that as x\to{0} , \sin(u(x))\sim{u(x)}

    so we could just say as x\to{0} , \sin(\sin(x))\sim{\sin(x)}

    so \lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}=\lim_{  x\to{0}}\frac{\sin(x)}{\sin(x)}=1
    Is it possible to simplify it by making t = sinx
    \lim_{x\to{0}}\frac{\sin(t)}{t}

    and solving that?
    That's what I did and came up with the answer, 1.
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  9. #9
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    for this one:
    \lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}
    can I apply L'Hopitals rule. Differentiate d/dx:

    \frac{-64x^{-1.4}}{-3.2x^{-1.4}}

    = \frac{64}{3.2}

    =20

    Das ist gut, ja?
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  10. #10
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    Hello,

    Quote Originally Posted by Dr Zoidburg View Post
    Is it possible to simplify it by making t = sinx
    \lim_{{\color{red}t}\to {0}}\frac{\sin(t)}{t}

    and solving that?
    That's what I did and came up with the answer, 1.
    Yeah
    (tiny mistake, in red)

    Quote Originally Posted by Dr Zoidburg View Post
    for this one:
    \lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}
    can I apply L'Hopitals rule. Differentiate d/dx:

    \frac{-64x^{-1.4}}{-3.2x^{-1.4}}

    = \frac{64}{3.2}

    =20

    Das ist gut, ja?
    Yop
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