Lim x - 01-2 x-1-3 x1-x2- e1-x-

X→ 0lime^1/x^2(log(1+2x) log(1+3x))+1/x e^x→ 0lim(log(1+2x) log(1+3x))+x/x^2 Using the expansion we get =e^5/2 ...

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Expansions to Remove Indeterminate Form

lim x → 01+2 ...

Question

$l i m x \to 0 {(\frac{1 + 2 x}{1 + 3 x})}^{\frac{1}{x^{2}}} . e^{\frac{1}{x}} =$

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