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OLga [1]
3 years ago
10

Please help me find limit​

Mathematics
1 answer:
cluponka [151]3 years ago
4 0

9514 1404 393

Answer:

  -13/11

Step-by-step explanation:

Straightforward evaluation of the expression at x=1 gives (1 -1)/(1 -1) = 0/0, an indeterminate form. So, L'Hopital's rule applies. The ratio of derivatives is ...

  \displaystyle\lim_{x\to 1}\dfrac{n}{d}=\dfrac{n'}{d'}=\left.\dfrac{\dfrac{4}{3\sqrt[3]{4x-3}}-\dfrac{7}{2\sqrt{7x-6}}}{\dfrac{5}{2\sqrt{5x-4}}-\dfrac{2}{3\sqrt[3]{2x-1}}}\right|_{x=1}=\dfrac{4/3-7/2}{5/2-2/3}=\dfrac{8-21}{15-4}\\\\=\boxed{-\dfrac{13}{11}}

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