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This is a question from an exam on calculus. $$ \lim_{x \rightarrow \infty }(-6e^{11x} + 9sin(x) + 3e^{8x} )^{8/4x} $$

I tried this:

$$ (3e^{(8x)})^{(8/4x)} < (-6e^{11x} + 9sin(x) + 3e^{8x} )^{8/4x} < (3e^{8x}+3e^{8x}+3e^{8x} )^{8/4x} $$

However, I'm not sure if it's right. The problem is: $ (3e^{(8x)})^{(8/4x)} < (-6e^{11x} + 9sin(x) + 3e^{8x} )^{8/4x} $

If it was $+6e$ instead of $-6e$,then I am sure that it's true, because it's make sense that the middle expression is bigger, because the middle expression is the same like the left side , only with more additions. but...

There is a minus on the middle expression, how can I know that $ (3e^{(8x)})^{(8/4x)} < (-6e^{11x} + 9sin(x) + 3e^{8x} )^{8/4x} $ is really true?

Can I use the squeeze theorem on this exercise?

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  • $\begingroup$ Do you need/want to use the squeeze theorem, or just want to compute the limit? There is IMHO a much faster way to compute it $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 14:43
  • $\begingroup$ sorry, but isn't $-6 e^{11x}+9\sin(x)+3 e^{8x}<0$ for large $x$? Then your limit is not properly defined... $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 15:00
  • $\begingroup$ @Pierpaolo Vivo I just want to compute the limit. I don't know what is mean limit not defined. you mean that there is no limit? this is a question from end of semester exam, and according to the exam solution there is limit , e^2 , which is properly wrong because in wolfram the solution is e^22 $\endgroup$ – Silas2033 Jan 24 '16 at 17:47
  • $\begingroup$ Take $x=5$. Your function $(3 \exp (8 x)-6 \exp (11 x)+9 \sin (x))^{2/x}$ computed for $x=5$ is $=2.26841\times 10^9+6.98145\times 10^9 \mathrm{i}$. It is a complex number. Your question is not conceptually different from asking: what is the limit $\lim_{x\to -5}\sqrt{x}$. $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 18:06
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Either your lecturer is rather cheeky, or there is a typo in your question. The function $$ h(x)=-6 e^{11x}+9\sin(x)+3 e^{8x} $$ is badly negative for $x\to +\infty$. Take the limit $x\to +\infty$. You can compute it easily as $$ e^{11x}\left(-6 +9\frac{\sin(x)}{e^{11x}}+3 e^{-3x}\right)\to -\infty $$ (where en passant you can also use the squeeze theorem to show that $\sin(x)/e^{11x}\to 0$ !). Therefore, your limit is not properly defined on the reals, as the domain of a function of the form $h(x)^{w(x)}$ is $h(x)>0$ (otherwise you cannot use the step $h(x)^{w(x)}=\exp(w(x)\ln h(x)$).

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  • $\begingroup$ there is no typo. I checked. the solution is e^2 . but he didn't use squeeze thorem $\endgroup$ – Silas2033 Jan 24 '16 at 17:12
  • $\begingroup$ Well, it is a fact that you cannot raise a negative function to another function within the reals. You can do it only using the principal value of a logarithm in the complex plane, which I doubt that is what you were requested to do during the exam. If you try to plot either the $h(x)$ of my answer, or your original function, you will see confirmation of my statement. You cannot even plot your function for large $x$! $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 17:26
  • $\begingroup$ the same function is in the exam answers and in the exam solution pdf file so I don't think that there is a 2 typos . I can send an email to the lecturer but can you please explain the problem in simple English so I can translate it? or please make it shorter and I send it as is $\endgroup$ – Silas2033 Jan 24 '16 at 18:08
  • $\begingroup$ I asked the lecturer this question in the class. you were right. the limit is not belong to "R" field. he was surprised that this question was in the exam. $\endgroup$ – Silas2033 Jan 27 '16 at 17:07
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$$\lim_{x\to\infty}\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)^{\frac{8}{4x}}=$$ $$\lim_{x\to\infty}\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)^{\frac{2}{x}}=$$ $$\lim_{x\to\infty}\exp\left[\ln\left(\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)^{\frac{2}{x}}\right)\right]=$$ $$\lim_{x\to\infty}\exp\left[\frac{2}{x}\ln\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)\right]=$$ $$\exp\left[\lim_{x\to\infty}\frac{2\ln\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)}{x}\right]=$$ $$\exp\left[\lim_{x\to\infty}\frac{\frac{\text{d}}{\text{d}x}\left(2\ln\left(-6e^{11x}+9\sin(x)+3e^{8x}\right)\right)}{\frac{\text{d}}{\text{d}x}\left(x\right)}\right]=$$ $$\exp\left[2\lim_{x\to\infty}\frac{\frac{8e^{8x}-22e^{11x}+3\cos(x)}{e^{8x}-2e^{11x}+3\sin(x)}}{1}\right]=$$ $$\exp\left[2\lim_{x\to\infty}\frac{8e^{8x}-22e^{11x}+3\cos(x)}{e^{8x}-2e^{11x}+3\sin(x)}\right]=$$ $$\exp[2\cdot11]=\exp[22]=e^{22}$$

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    $\begingroup$ can you confirm that $-6 e^{11x}+9\sin(x)+3 e^{8x}<0$ for large $x$? If yes, the function is not properly defined, right? [we cannot take the \exp(\log(.))] $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 15:03

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