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Math Help - [SOLVED] Exponential Limits

  1. #1
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    [SOLVED] Exponential Limits

    Hi guys

    I'm at the stage of my calculus course where we've just been introduced to the logarithm and exponential functions and I'm having some trouble evaluating limits in the form of exponential functions.

    lim as x->0 (exp(2x)+2x)^1/x, so far i've gotten up to lim x->0 (1+2x/exp(2x))^1/x*e^2

    lim as x->0 (((1+x+x^2)^1/x)/exp(1))^1/x

    I know that the number e can be defined as the limit as x goes to 0 of (1+x)^1/x, but I'm at a complete loss when it comes to applying this to evaluating the given limits.

    Any help would be appreciated.

    Thanks.
    Last edited by Hweengee; October 11th 2008 at 12:20 AM.
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  2. #2
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    Here is a quicky way, so to speak. We can do it without, but what the heck.

    \lim_{x\to 0}\left(e^{2x}+2x\right)^{\frac{1}{x}}

    Let t=2x:

    \lim_{t\to 0}\left(e^{t}+t\right)^{\frac{2}{t}}

    Rewrite:

    \displaystyle{^{2\lim_{t\to 0}\frac{ln(e^{t}+t)}{t}}}

    Now, we'll take the easy way out and use L'Hopital, giving:

    \displaystyle{e^{2\lim_{t\to 0}\frac{e^{t}+1}{e^{t}+t}}}

    Now, the limit can be seen by merely plugging in t=0.

    We could also do it without the t substitution by rewriting it as:

    e^{\lim_{x\to 0}\frac{ln(e^{2x}+2x)}{x}}
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  3. #3
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    No need L'H˘pital, note that,

    \begin{aligned}<br />
   \left( e^{2x}+2x \right)^{1/x}&=\exp \left\{ \frac{\ln \left( e^{2x}+2x \right)}{x} \right\} \\ <br />
 & =\exp \left\{ \frac{2x+\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{x} \right\} \\ <br />
 & =\exp \left\{ 2+\frac{\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{\dfrac{2x}{e^{2x}}}\cdot \frac{2}{e^{2x}} \right\},<br />
\end{aligned}

    hence this last expression is e^4 as x\to0 and we're done.
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  4. #4
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    No need L'H˘pital
    I know K-meister. I was just lazy.

    It's always nice to do them without if possible. I wonder why we dissuade ourselves from using the Hospital rule?.

    It makes things easier. It's just a thing I suppose. I like to veer away from it, too, most of the time.
    Last edited by galactus; October 11th 2008 at 08:38 AM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    \left(e^{2x}+2x\right)^{\frac{1}{x}}=e^{\frac{\ln(  e^{2x}+2x)}{x}}\sim{e^{\frac{\ln(1+2x+2x)}{x}}}=e^  {\frac{\ln(1+4x)}{x}}\sim{e^{\frac{4x}{x}}}=e^{4}
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  6. #6
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \left(e^{2x}+2x\right)^{\frac{1}{x}}=e^{\frac{\ln(  e^{2x}+2x)}{x}}\sim{e^{\frac{\ln(1+2x+2x)}{x}}}=e^  {\frac{\ln(1+4x)}{x}}\sim{e^{\frac{4x}{x}}}=e^{4}
    How do you get from step 2 to step 3?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    How do you get from step 2 to step 3?
    Do you know about asymptotic equivalences?
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  8. #8
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Do you know about asymptotic equivalences?
    No.......? I'm halfway through Calculus II.
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  9. #9
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    Quote Originally Posted by Krizalid View Post
    No need L'H˘pital, note that,

    \begin{aligned}<br />
exp \left\{ 2+\frac{\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{\dfrac{2x}{e^{2x}}}\cdot \frac{2}{e^{2x}} \right\},<br />
\end{aligned}

    hence this last expression is e^4 as x\to0 and we're done.
    sorry, i still dont get this last bit. how does the expression become exp(4) when x goes to 0? doesn't the denominator of the expression on the right become 0 as x goes to 0? so how does the expression on the right become 2 as x goes to 0?
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  10. #10
    Moo
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    Hello,
    Quote Originally Posted by Hweengee View Post
    sorry, i still dont get this last bit. how does the expression become exp(4) when x goes to 0? doesn't the denominator of the expression on the right become 0 as x goes to 0? so how does the expression on the right become 2 as x goes to 0?
    \exp(0)=e^0=1

    So \frac{2}{e^{2x}} \stackrel{x \to 0}{\longrightarrow} \frac 21=2


    Moreover, \dfrac{\ln \left(1+\tfrac{2x}{e^{2x}}\right)}{\tfrac{2x}{e^{2  x}}} \stackrel{x \to 0}{\longrightarrow} 1, because if you make a substitution t=\frac{2x}{e^{2x}}, you'll have :

    \frac{\ln(1+t)}{t}, which tends to 1 as t tends to 0.
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  11. #11
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    Quote Originally Posted by Moo View Post
    \frac{\ln(1+t)}{t}, which tends to 1 as t tends to 0.
    Hi, is this bit by definition? since the number e is defined as the limit of (1+x)^1/x? and ln e is 1? Thanks for showing me how it works out, I was really lost before this.
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  12. #12
    Moo
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    Quote Originally Posted by Hweengee View Post
    Hi, is this bit by definition? since the number e is defined as the limit of (1+x)^1/x? and ln e is 1? Thanks for showing me how it works out, I was really lost before this.
    Hmmm here is my method =)

    I'll use this definition : f'(a)=\lim_{t \to 0} ~ \frac{f(a+t)-f(a)}{t}

    Now, let f(x)=\ln(1+x)
    You can see that f(0)=\ln(1+0)=\ln(1)=0

    So \lim_{t \to 0} ~ \frac{\ln(1+t)}{t}=\lim_{t \to 0} ~ \frac{\ln(1+t)-0}{t}=\lim_{t \to 0} ~ \frac{\ln(1+t)-\ln(1)}{t}=\lim_{t \to 0} ~ \frac{f(t)-f(0)}{t}

    =f'(0)=\left(\ln(1+t)\right)'(0)=\frac{1}{1+0}=1


    Otherwise, you can use l'Hospital's rule
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  13. #13
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    Hi

    I think I've gotten the 1st limit worked out, thanks to moo and the rest, but for the second I'm stuck at

    \lim_{x\to 0}~{\frac {\ln \left( 1+x+{x}^{2} \right) -x}{{x}^{2}}}

    Any suggestions on how to carry on? Or should I just use L'Hopital's rule?
    Last edited by Hweengee; October 12th 2008 at 04:37 AM.
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  14. #14
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    You could try using the series expansion:

    ln(1+x+x^{2})=x+\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{3}+..........

    Subtract off the x and we are left with:

    ln(1+x+x^{2})=\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{3}+.........

    Now, divide by x^2 and get:

    \frac{1}{2}-\frac{2x}{3}+\frac{x^{2}}{4}+\frac{x^{3}}{5}-\frac{x^{4}}{3}+...........

    Now, take the limit as x-->0. See what it is?. Easy to see now.
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  15. #15
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    yup. thanks for the help, much appreciated.
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