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I need to solve this limit:

$$\lim\limits_{x\rightarrow +\infty }\frac{\left[ 2-e^{\frac{1}{x}} +\sin\left(\frac{1}{x}\right) -\cos\left(\frac{1}{x}\right)\right]^{2}}{\frac{1}{3}\left(\frac{1}{x}\right)^{6}}$$

I tried using the substitution $t=\frac1x$ so that the limit becomes

$$\lim\limits_{t\rightarrow 0}\frac{\left[ 2-e^{t} +\sin( t) -\cos( t)\right]^{2}}{\frac{1}{3}( t)^{6}}$$

Then I tried to trace it back to the notable limits of the exponential, the sine and the cosine by dividing the 2 into two 1s

$$3\lim\limits _{t\rightarrow 0}\frac{\left[ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)\right]^{2}}{t^{6}}$$

but I don't know how to distribute the denominator and how to handle the square of the numerator. I tried solving the square and use l'Hopitals rule but it really doesn't get me anywhere. I'm sure i'm missing something simple, but can't really figure out what. Thanks in advance to anyone who decides to help me.

Sine of the Time
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W. White
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  • At first you can write $t^6=(t^3)^2$ and write the whole fraction as a square. – Lereu Oct 28 '23 at 12:45
  • and then? I still have the problem that I can't distribute the denominator as I would like because it is a sum and not a product :(. Also I think the answer is 1/3, maybe it is of any help. – W. White Oct 28 '23 at 13:09

2 Answers2

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Using the following Taylor polynomials of order $3$ at $0$ $$e^t = 1 + t + \frac{t^2}2 + \frac{t^3}6 + o(t^3)$$ $$\sin(t) = t - \frac{t^3}6 + o(t^3)$$ $$\cos(t) = 1 - \frac{t^2}2 + o(t^3)$$ we obtain $$\lim_{t\to 0}\frac{2- e^t+\sin t-\cos t}{t^3} = \lim_{t\to 0}\frac{2-(1+t+\frac{t^2}2+\frac{t^3}6)+(t-\frac{t^3}6)-(1-\frac{t^2}2)+o(t^3)}{t^3}= \lim_{t\to 0}\frac{2-1-t- \frac{t^2}2-\frac{t^3}6+t-\frac{t^3}6-1+\frac{t^2}2+o(t^3)}{t^3}=-\frac13$$ Plug it into what you already have and you get that the original limit equals $1/3$.

Roman Hric
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    A thousand thanks! ☺ I didn't think about using Taylor. Just out of curiosity, do you think there is a way to solve this with famous limits? The exercise almost seems to be made for that – W. White Oct 28 '23 at 13:36
  • I guess a clever manipulation would transform it to a combination of known limits. – Roman Hric Oct 28 '23 at 16:24
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You mentioned that you used L'Hospital too. It works. You have $$ f(t) = 3\lim\limits_{t\rightarrow 0}\frac{\left[ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)\right]^{2}}{t^{6}}. $$ It is $$ \frac{\left[ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)\right]^{2}}{t^{6}} = \frac{\left[ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)\right]^{2}}{(t^{3})^2} =$$$$ \left(\frac{ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)}{t^{3}}\right)^2. $$ Let $$ g(t) = \frac{ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)}{t^{3}}. $$ Then (using L'Hospital where I write an arrow) $$ \frac{ -\left( e^{t} -1\right) +\sin( t) +1-\cos( t)}{t^{3}} \stackrel{t\rightarrow 0}{\rightarrow} \frac{ -e^t + \cos(t) + \sin(t)}{3t^{2}} \stackrel{t\rightarrow 0}{\rightarrow} \frac{ -e^t -\sin(t)+\cos(t)}{6t^{1}}$$$$ \stackrel{t\rightarrow 0}{\rightarrow} \frac{ -e^t - \cos(t) - \sin(t)}{6} \stackrel{t\rightarrow 0}{\rightarrow} -\frac{1}{3}. $$

We obtain our solution, hence $$ \lim\limits_{t\rightarrow 0} f(t) = 3 \cdot (-\frac{1}{3})^2 = \frac{1}{3}. $$

Sebastiano
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Lereu
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