lolerde4
contestada

Find the exact value of the following limit: the limit as x goes to 0 of the quotient of the quantity e raised to the 3 times x power minus 3 times x minus 1 and x squared.
Type answer as a decimal.

Respuesta :

konrad509
[tex]\displaystyle \lim_{x\to 0}\dfrac{e^{3x}-3x-1}{x^2}=\\ \lim_{x\to 0}\dfrac{(e^{3x}-3x-1)'}{(x^2)'}=\\ \lim_{x\to 0}\dfrac{3e^{3x}-3}{2x}=\\ \lim_{x\to 0}\dfrac{(3e^{3x}-3)'}{(2x)'}=\\ \lim_{x\to 0}\dfrac{9e^{3x}}{2}=\\ \dfrac{9e^{3\cdot0}}{2}=\dfrac{9}{2}=4.5 [/tex]

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