Sunday, 11 May 2014

calculus - How to solve limxtosqrt2left(fracex2+e2(1x2)[ln(x23sqrt2x+5)]2right)?



I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
lim



Answer



\begin{aligned} \lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2\left(1-x^2\right)}{\left(ln\left(x^2-3\sqrt{2}x+5\right)\right)^2}\right) & = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(1-\left(t+\sqrt{2}\right)^2\right)}{\left(ln\left(\left(t+\sqrt{2}\right)^2-3\sqrt{2}\left(t+\sqrt{2}\right)+5\right)\right)^2}\right) \\& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(-t^2-2\sqrt{2}t-1\right)}{\ln \:^2\left(t^2-\sqrt{2}t+1\right)}\right) \\& = \lim _{t\to 0}\left(\frac{e^2+2\sqrt{2}e^2t+5e^2t^2+o\left(t^2\right)+e^2\left(-t^2-2\sqrt{2}t-1\right)}{2t^2+o\left(t^2\right)}\right) \\& = \color{red}{2e^2} \end{aligned}

Solved with substitution t = x-\sqrt2 and Taylor expansion


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