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I was assisting a TA for an introductory calculus class with the following limit,

$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x}$$

and I came to simple solution which involved evaluating the "reciprocal" limit

$$\lim_{z \rightarrow 0} \left(\frac{1}{1+\frac{z}{5}}\right)^{8/z}$$

by using the Taylor expansion of $\log(1+z)$ around $z=0$. However, the TA claims that the students have not learned about series expansions so that would not be a valid solution for the course. I tried applying L'Hopital's rule, which I was told the class did cover, but I was unsuccessful. As a note I will mention that

$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{-8/5}.$$

Any ideas for a solution to this problem using only knowledge from a first quarter (or semester) calculus course which hasn't covered series expansions?

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    $\begingroup$ So what is $e$ for these students? $\endgroup$ – Alex B. Nov 26 '10 at 7:05
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    $\begingroup$ @Alex Troesch: a general strategy for evaluating indeterminate forms of type $0^0$, $1^{\infty}$, and $\infty^0$ is to use the identity $a^b = e^{b\ln a}$, and then use L'Hopital's Rule on the exponent (after suitable rewriting). $\endgroup$ – Arturo Magidin Dec 8 '10 at 3:11
  • $\begingroup$ And the "Most useless edit 2011 award" goes to... Max/Foool!!!!!!! $\endgroup$ – The Chaz 2.0 Mar 22 '12 at 16:32
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If they know the definition of $e$ as $$\lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n,$$ then set $\alpha=\lim_{x\rightarrow\infty}\left(\frac{20x}{20x+4}\right)^{8x}$ and note that $$1/\alpha^5 = \lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{40x}= \left(\lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{5x}\right)^8 = e^8.$$

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  • $\begingroup$ Considering $1/\alpha^5$ I don't think is a natural choice for the students, but I think this answer is clear. Hopefully they have covered that 'definition' of $e$. You can never be sure with these intro courses sometimes! $\endgroup$ – Alex Troesch Nov 26 '10 at 7:28
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    $\begingroup$ Here is the motivation, so that it seems more natural to them: considering 1/alpha should seem natural to them, since alpha has a sum in the denominator and only one term (matching one of the summands) in the numerator. They should feel the urge to take the -1st power to split the sum apart. Next, taking the 5-th power might not seem natural until the penultimate step, when they realise that there is a 1/5x floating around. If they look at the definition for e, they should get annoyed by the 5 and feel the urge to come closer to the actual definition of e, so taking 5th power becomes natural. $\endgroup$ – Alex B. Nov 26 '10 at 7:34
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HINT:

Always start by simplifying $$\left(\frac{20x}{20x+4}\right)^{8x}=\left(\frac{20x+4}{20x}\right)^{-8x}=\left(1+\frac{1}{5x}\right)^{-8x}$$ perhaps a substitution helps...

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HINT:

\begin{align*} \lim_{x \to \infty} \biggl(\frac{20x}{20x+4}\biggr)^{8x} &=\lim_{x \to \infty} \biggl( 1 - \frac{4}{20x+4}\biggr)^{8x} \\ &=\lim_{x \to \infty} \biggl(1 - \frac{4}{20x+4}\biggr)^{\frac{20x+4}{4} \cdot \frac{8x}{20x+4} \cdot 4}&\end{align*}

Now use the fact that $$\lim_{n \to \infty} \biggl(1 + \frac{1}{n}\biggr)^{n} =e$$

ADDED: After simplyfing you should get this. $$e^{\displaystyle \lim_{x \to \infty} - \frac{8x}{20x+4} \cdot 4} = e^{-8/5}$$

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Hint: Consider taking logarithms before applying L'Hospital's rule.

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We can calculate Ln() of the function and result is exp() of the answer. $$ \lim_{x \rightarrow \infty} \left[ ( \frac{20x}{20x+4})^{8x} \right] = \lim_{x \rightarrow \infty} \left[ ( \frac{5x}{5x+1})^{8x} \right]$$ $$ Ans= \lim_{x \rightarrow \infty} \left[ 8x. Ln ( \frac{5x}{5x+1}) \right] = \infty . 0 $$ $$ Ans=8 \lim_{x \rightarrow \infty} \left[ \frac{ Ln ( \frac{5x}{5x+1})}{\frac{1}{x}} \right] = \frac{0}{0} \space \rightarrow \space Ambiguity in Mathematics $$ we can use Hopital: $$ Hopital : \frac{derivative \space of \space the \space Numerator}{derivative \space of \space the \space denominator}$$ $$ Ans=8\lim_{n \rightarrow \infty}\left[ \frac{ \frac{1}{x(5x+1)}}{\frac{-1}{x^2}} \right]=8\lim_{n \rightarrow \infty} \left[ \frac{-x}{5x+1} \right]=8(\frac{-1}{5}) $$ therefor : $$ \lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{Ans}=e^{\frac{-8}{5}} $$

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