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.
Best Answer
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 $1/3$.