Find the exact value of the following limit: lim x goes to 0 (e^8x-8x-1)/x^2

Question

Find the exact value of the following limit: lim x goes to 0 (e^8x-8x-1)/x^2

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Hưng Khoa 5 months 2021-08-29T12:30:00+00:00 1 Answers 2 views 0

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    2021-08-29T12:31:31+00:00

    Answer:

    32

    Step-by-step explanation:

    We can apple L’Hopital’s Rule to solve this problem:

    \lim _{x\to \:\:0}\left(\frac{e^{8x}-8x-1}{x^2}\right)\\=> \lim _{x\to \:0}\left(\frac{8e^{8x}-8}{2x}\right)\\\\=> \lim _{x\to \:0}\left(\frac{4\left(e^{8x}-1\right)}{x}\right)\\\\=> \lim _{x\to \:0}\left(\frac{32e^{8x}}{1}\right)\\\\=>\frac{32e^{8\cdot \:0}}{1}\\\\=> 32

    Hope this helps!

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