My try
But after that I stuck . Can anbody help me .
Please explain me as I am a high school student
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1Use Hopital method! – Ken Apr 27 '17 at 02:36
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Related: A similar but not identical limit. – Frenzy Li Apr 27 '17 at 07:04
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@FrenzyLi thanks it is very useful – search Apr 27 '17 at 07:13
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Note that we can write
$$\begin{align} (1+3t\pm 2t^2)^{1/t}&=e^{\frac1t \log(1+3t\pm 2t^2)}\\\\ &=e^{3\pm 2t -\frac92 t+O(t^2)}\\\\ &=e^3\left(1+\pm 2t-\frac92 t+O(t^2)\right) \end{align}$$
Thus, the limit of interest is
$$\lim_{t\to 0}\left(\frac{(1+3t+ 2t^2)^{1/t}-(1+3t- 2t^2)^{1/t}}{t}\right)=e^3\lim_{t\to0}(4+O(t))=4e^3$$
Mark Viola
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Note that $\log(1+x)=x+O(x^2)$. So, with $x=3t\pm 2t^2$ we have$$\frac1t \log(1+3t\pm 2t^2)=\frac1t \left((3t\pm 2t^2 )+O(3t\pm 2t^2)^2\right)=3+O(t)$$ – Mark Viola Apr 27 '17 at 03:14
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@Displayname Yes, I did miss that and have edited to correct thanks to your pointing that out! Much appreciative! -Mark – Mark Viola Apr 27 '17 at 06:01