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Find 
lim_(h rarr0)(5log(2+h)-5log(2))/(h).
Choose 1 answer:
(A) 
(5)/(2)
(B) 
(5)/(2ln(10))
(C) 
5log(2)
(D) The limit doesn't exist

Find limh05log(2+h)5log(2)h \lim _{h \rightarrow 0} \frac{5 \log (2+h)-5 \log (2)}{h} .\newlineChoose 11 answer:\newline(A) 52 \frac{5}{2} \newline(B) 52ln(10) \frac{5}{2 \ln (10)} \newline(C) 5log(2) 5 \log (2) \newline(D) The limit doesn't exist

Full solution

Q. Find limh05log(2+h)5log(2)h \lim _{h \rightarrow 0} \frac{5 \log (2+h)-5 \log (2)}{h} .\newlineChoose 11 answer:\newline(A) 52 \frac{5}{2} \newline(B) 52ln(10) \frac{5}{2 \ln (10)} \newline(C) 5log(2) 5 \log (2) \newline(D) The limit doesn't exist
  1. Combine logarithmic terms: Use the properties of logarithms to combine the terms in the numerator. limh05log(2+h)5log(2)h=limh05log(2+h2)h\lim_{h \rightarrow 0}\frac{5\log(2+h)-5\log(2)}{h} = \lim_{h \rightarrow 0}\frac{5\log\left(\frac{2+h}{2}\right)}{h}
  2. Apply constant multiple rule: Apply the constant multiple rule in limits to take the constant 55 out of the limit.limh0(5log(2+h2)h)=5×limh0(log(2+h2)h)\lim_{h \to 0}\left(\frac{5\log\left(\frac{2+h}{2}\right)}{h}\right) = 5 \times \lim_{h \to 0}\left(\frac{\log\left(\frac{2+h}{2}\right)}{h}\right)
  3. Use derivative definition: Use the definition of the derivative for the function f(x)=log(x)f(x) = \log(x) at x=2x = 2.limh0log((2+h)/2)h=f(2)\lim_{h \to 0}\frac{\log((2+h)/2)}{h} = f'(2) where f(x)=log(x)f(x) = \log(x)
  4. Calculate derivative: The derivative of log(x)\log(x) is 1xln(10)\frac{1}{x\ln(10)}. So, f(2)=12ln(10)f'(2) = \frac{1}{2\ln(10)}.
  5. Multiply by constant: Multiply the derivative by the constant 55 that we factored out earlier.\newline5×f(2)=5×(12ln(10))=52ln(10)5 \times f'(2) = 5 \times \left(\frac{1}{2\ln(10)}\right) = \frac{5}{2\ln(10)}

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