Question

limit X tends to 0 5^x-4^x/3^x-1
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Answer 1

Answer:$\frac{ln 5 - ln 4}{ln 3}$

Step-by-step explanation:

$\lim_{x\to 0} \frac{5^{x} -4^{x} }{3^{x} -1} = \frac{5^{0} -4^{0} }{3^{0} -1} \\ \\ =\frac{1-1}{1-1} =\frac{0}{0}=0$ (not possible)

∴By using L'hospital rule

$\lim_{x\to 0} \frac{5^{x} -4^{x} }{3^{x} -1}=[tex]\lim_{x \to 0} \frac{5^{x}  ln 5 - 4^{x} ln 4 }{3^{x} ln 3 -0 }$$\\ \\  =\frac{5^{0} ln 5 - 4^{0} ln 4  }{3^{0} ln 3} =\frac{ln 5 - ln 4}{ln 3}$