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Find $$ \lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})} {\log_2( 1+x^{-2} + x^{-3})}$$

Firstly I replace $x$ with $1/t$. Then $t$ tends to $0$. And then I fix numerator with formulas for limits considering $t \to 0$. But I don't know what to do with denominator. Any help?

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    I've improved your question's formatting; apologies if I changed your meaning. You can [see here](http://math.stackexchange.com/posts/2133726/revisions) how I edited your question. [Please see here](http://meta.math.stackexchange.com/q/5020/264) for a guide to writing math with MathJax, and [see here](http://math.stackexchange.com/help/formatting) for a guide to formatting posts with Markdown. – 2017-02-07
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    Sqrt is just over 1 - cos ... not over the whole nominator . idk how to fix it – 2017-02-07
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    How's the edited version? – 2017-02-07
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    Fixed :) thanks – 2017-02-07

4 Answers 4

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Since it is the denominator term that is the source of concern, note that after the substitution $x=1/t$, we have

$$\begin{align} \log_2(1+x^{-2}+x^{-3})&=\log_2(1+t^2+t^3)\\\\ &=\left(\frac{\log_2(1+t^2+t^3)}{t^2+t^3}\right)\,(t^2+t^3)\\\\ &=\left(\frac{\log(1+t^2+t^3)}{\log(2)\,(t^2+t^3)}\right)\,(t^2+t^3) \end{align}$$

For the numerator, we see that

$$\begin{align} \sqrt{1-\cos^3(1/x)}&=\sqrt{1-\cos^3(t)}\\\\ &=\sqrt{1-\cos(t)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\ &=\sqrt{2\sin^2(t/2)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\ &=\sqrt{2}\sin(t/2)\sqrt{1+\cos(t)+\cos^2(t)} \end{align}$$

Additionally, we have

$3^{1/x}-5^{-1/x}=3^t\left(1-e^{-\log(15)t}\right)$$

Now, we can write

$$\begin{align} \frac{\sqrt{1-\cos^3(1/x)}\left(3^{1/x}-5^{-1/x}\right)}{\log_2(1+x^{-2}+x^{-3})}&=\frac{\sqrt{1-\cos^3(t)}\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\ &=\sqrt{2}\sqrt{1+\cos(t)+\cos^2(t)}\frac{\sin(t/2)\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\ &=\color{blue}{\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}}\frac{\color{red}{\sin(t/2)}\color{green}{\left(1-e^{-\log(15)t}\right)}}{\color{purple}{\log_2(1+t^{2}+t^{3})}}\\\\ &=\color{blue}{\left(\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}\right)}\\\\ &\times \frac{\color{red}{\sin(t/2)}}{\color{orange}{t/2}}\\\\ &\times \frac{\color{gold}{t^2+t^3}}{\color{purple}{\frac{\log(1+t^2+t^3)}{\log(2)}}}\\\\ &\times \frac{\color{green}{1-e^{-\log(15)t}}}{\log(15)t}\\\\ &\times \color{orange}{\frac{t}{2}}\times {\log(15)t}\times\frac{1}{\color{gold}{t^2+t^3}} \end{align}$$

Can you finish now?

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    Oh right ! I stucked right at that end where you got 1/(t^2+t^3) * t^2 . I really thought that expression tends to 0 but after I wrote it down i realized its acctually 1. Haha , i feel so dumb now ... Thanks. – 2017-02-07
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    You're welcome! My pleasure. -Mark – 2017-02-07
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    Tried but it says I need to have reputation above 15 ... ? – 2017-02-07
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    @JamesGroon Now you can upvote. try again :). – 2017-02-07
1

$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})} {\log_2( 1+x^{-2} + x^{-3})}$$

$$\log_2( 1+x^{-2} + x^{-3}) = {\ln (1+x^{-2} + x^{-3})\over \ln 2}$$ let $t = x^{-2} + x^{-3}$ $${\ln (1+t) \over t\ln 2} \overbrace{\large{=}} ^{ t \to 0} {1 \over \ln 2} \tag {1}$$.

Using (1) in our question, $$\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})} {\log_2( 1+x^{-2} + x^{-3})} = (\ln 2)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})} {x^{-2} + x^{-3}}\tag{1*}$$

$${3^{1/x} - {{1\over 5^{1/x}}}\over (x^{-1})} = {15^{1/x} - 1\over5^{1/x}(x^{-1})} \overbrace{ \Large{=}}^{x\to \infty} \ln(15) \tag{2}$$

using (2) in (1*),

$$(\ln 2)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})} {x^{-2} + x^{-3}} = (\ln2)(\ln15)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}}{x^{-1} + x^{-2}}\tag{2*}$$

let t = 1/x

$${\left(\sqrt{1-\cos^3 t}\right)^\prime\over (t + t^2)^\prime} = {3\sin t \times \cos^2 t\over (1 + 2t)\times 2 \times \sqrt{1-\cos^3 t}} = {3\sin t \times \cos^2 t\over (1 + 2t)\times 2 \times \sqrt{2}\sin(t/2)\sqrt{1+\cos(t)+\cos^2(t)}} = {3(\cos t/2) \times \cos^2 x\over (1 + 2t)\times \sqrt{2}\sqrt{1+\cos(t)+\cos^2(t)}} \overbrace{\Large{=}}^{t \to 0} {3 \over \sqrt 2}$$ $$\tag{3}$$

using (3) in (2*),

$$(\ln2)(\ln15)\lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}}{x^{-1} + x^{-2}} = {3 \over \sqrt 2}\ln 2\ln 15 $$

0

$$\lim_{t\to 0}\frac{\sqrt{1-\cos ^3\left(t\right)}\cdot \left(3^t-5^{-t}\right)}{\log _2\left(\:1+t^2\:+\:t^3\right)}$$

Using Taylor approximation:

$$= \lim _{t\to 0}\left(\frac{\left(\sqrt{\frac{3}{2}}t+o(t)\right)\cdot \left(ln\left(15\right)t+o(t)\right)}{\frac{1}{\ln \left(2\right)}t^2+o(t^2)}\right) = \color{red}{\sqrt{\frac{3}{2}}\ln \left(2\right)\ln \left(15\right)}$$

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    Confirmed using *Mathematica*. – 2017-02-07
0

$$\lim_{t\to0^+}\frac{\sqrt{1-\cos(t)^3}\cdot\left(3^t-\frac{1}{5^t}\right)}{\frac{\ln(1+t^2+t^3)}{\ln(2)}}=\ln(2)\cdot\lim_{t\to0^+}\frac{\sqrt{1-\cos(t)^3}\cdot\left(3^t-\frac{1}{5^t}\right)}{\ln(1+t^2+t^3)}$$

Now, let's consider instead the square of the expression inside the limit. Notice that for positive $t$ near $0$, the expression above is always positive, so we find a limit for the square, we take the positive square root of the limit and we're good.

$$\lim_{t\to0^+}\frac{\left(1-{\cos(t)}^3\right)\cdot{\left(3^t-\frac{1}{5^t}\right)}^2 }{{\ln(1+t^2+t^3)}^2}$$

We have an inderteminate of type $0/0$, so we may apply L'Hopital. In fact, we'll need to apply it several times. After four applications you'll find that the numerator is

$$8\, \left( 1- \left( \cos \left( t \right) \right) ^{3} \right) \left( {3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{ {5}^{t}}} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{3}+{\frac { \left( \ln \left( 5 \right) \right) ^{3}}{{5} ^{t}}} \right) +6\, \left( 1- \left( \cos \left( t \right) \right) ^{ 3} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{2}- {\frac { \left( \ln \left( 5 \right) \right) ^{2}}{{5}^{t}}} \right) ^{2}+48\, \left( \sin \left( t \right) \right) ^{3} \left( { 3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{{5}^{t}}} \right) + 2\, \left( 1- \left( \cos \left( t \right) \right) ^{3} \right) \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{4}-{\frac { \left( \ln \left( 5 \right) \right) ^{4}}{{5}^{t}}} \right) +36\, \left( \cos \left( t \right) \right) ^{3} \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{2}-{\frac { \left( \ln \left( 5 \right) \right) ^{2}}{{5} ^{t}}} \right) +36\, \left( \cos \left( t \right) \right) ^{3} \left( {3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{ {5}^{t}}} \right) ^{2}+60\,\cos \left( t \right) \left( \sin \left( t \right) \right) ^{2} \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) ^{2}-21\, \left( \cos \left( t \right) \right) ^{3} \left( { 3}^{t}- \left( {5}^{t} \right) ^{-1} \right) ^{2}-168\, \left( \cos \left( t \right) \right) ^{2}\sin \left( t \right) \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{{5}^{t}}} \right) -72\,\cos \left( t \right) \left( \sin \left( t \right) \right) ^{2} \left( { 3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{{5}^{t}}} \right) ^{2}-72\,\cos \left( t \right) \left( \sin \left( t \right) \right) ^{2} \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{2}-{\frac { \left( \ln \left( 5 \right) \right) ^{2}}{{5}^{t}}} \right) +72\, \left( \cos \left( t \right) \right) ^{2}\sin \left( t \right) \left( {3}^{t}\ln \left( 3 \right) +{\frac {\ln \left( 5 \right) }{ {5}^{t}}} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{2}-{\frac { \left( \ln \left( 5 \right) \right) ^{2}}{{5} ^{t}}} \right) +24\, \left( \cos \left( t \right) \right) ^{2}\sin \left( t \right) \left( {3}^{t}- \left( {5}^{t} \right) ^{-1} \right) \left( {3}^{t} \left( \ln \left( 3 \right) \right) ^{3}+{ \frac { \left( \ln \left( 5 \right) \right) ^{3}}{{5}^{t}}} \right)$$

with limit as $\to0^+$ given by $$72\ln(5)\ln(3)+36\ln(3)^2+36\ln(5)^2={\Big(6\big(\ln(3)+\ln(5)\big)\Big)}^2={\big(6\ln(15)\big)}^2.$$

The denominator, on the other hand, will be

$$6\,{\frac { \left( 6\,t+2 \right) ^{2}}{ \left( {t}^{3}+{t}^{2}+1 \right) ^{2}}}-36\,{\frac { \left( 3\,{t}^{2}+2\,t \right) ^{2} \left( 6\,t+2 \right) }{ \left( {t}^{3}+{t}^{2}+1 \right) ^{3}}}+48\, {\frac {3\,{t}^{2}+2\,t}{ \left( {t}^{3}+{t}^{2}+1 \right) ^{2}}}+22\, {\frac { \left( 3\,{t}^{2}+2\,t \right) ^{4}}{ \left( {t}^{3}+{t}^{2}+ 1 \right) ^{4}}}-48\,{\frac {\ln \left( {t}^{3}+{t}^{2}+1 \right) \left( 3\,{t}^{2}+2\,t \right) }{ \left( {t}^{3}+{t}^{2}+1 \right) ^{ 2}}}+24\,{\frac {\ln \left( {t}^{3}+{t}^{2}+1 \right) \left( 6\,t+2 \right) \left( 3\,{t}^{2}+2\,t \right) ^{2}}{ \left( {t}^{3}+{t}^{2} +1 \right) ^{3}}}-6\,{\frac {\ln \left( {t}^{3}+{t}^{2}+1 \right) \left( 6\,t+2 \right) ^{2}}{ \left( {t}^{3}+{t}^{2}+1 \right) ^{2}}}- 12\,{\frac {\ln \left( {t}^{3}+{t}^{2}+1 \right) \left( 3\,{t}^{2}+2 \,t \right) ^{4}}{ \left( {t}^{3}+{t}^{2}+1 \right) ^{4}}} $$

with limit as $\to0^+$ given by $24=(2\sqrt{6})^2$.

Hence, the limit of the squared expression as a whole is

$${\left(\frac{6\ln(15)}{2\sqrt{6}}\right)}^2={\left(\frac{\sqrt{6}}2\cdot\ln(15)\right)}^2$$

It follows that the limit of the initial expression, and answer to the question, is

$$\ln(2)\cdot\frac{\sqrt{6}}2\cdot\ln(15)=\frac{\sqrt{6}}2\ln(2)\ln(15)$$