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3 years ago

What is lim x → 0 e^2x - 1/ e^x - 1

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Hello, please consider the following.

$\displaystyle \forall x \in \mathbb{R}\\\\\dfrac{e^{2x}-1}{e^x-1}\\\\=\dfrac{(e^x)^2-1^2}{e^x-1}\\\\=\dfrac{(e^x-1)(e^x+1)}{e^x-1}\\\\=e^x+1\\\\\text{So, we can find the limit.}\\\\\lim_{x\rightarrow 0} \ {\dfrac{e^{2x}-1}{e^x-1}}\\\\=\lim_{x\rightarrow 0} \ {e^x+1}\\\\=e^0+1\\\\\large \boxed{\sf \bf \ =2 \ }$

Thank you

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2 years ago

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Step-by-step explanation:

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F(x,y,z) = (xy, yz, zx)

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$\bf G(x,y) = (x, y, 3-x^2-y^2)$

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we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

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