Find lim_(h rarr0)(9*2^(1+h)-9*2^(1))/(h) Choose 1 answer: (A) 9 (B) 18 (C) 18 ln(2) (D) The limit doesn't exist

Factor out common factor: Rewrite the expression by factoring out the common factor 9*2^1 from the numerator. So, lim_(h -> 0)(9*2^(1+h)-9*2^(1))/h = lim_(h -> 0)9*2^1(2^h-1)/h.Simplify expression: Simplify the expression by taking 9*2^1 out of the limit since it's a constant. So, 9*2^1 * lim_(h -> 0)(2^h-1)/h.Calculate constant: Calculate 9*2^1 which is 9*2 = 18. So, 18 * lim_(h -> 0)(2^h-1)/h.Use derivative definition: Use the definition of the derivative for the function f(x) = 2^x at x = 1. The limit becomes 18 * f'(1) where f'(x) = ln(2)*2^x.Calculate final result: Calculate f'(1) which is ln(2)*2^1 = ln(2)*2. So, 18 * ln(2)*2.

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Math Problems

Algebra 1

Multiply and divide rational expressions

Full solution

Q. Find

\lim _{h \rightarrow 0} \frac{9 \cdot 2^{1+h}-9 \cdot 2^{1}}{h}

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Choose

1

answer:

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(A)

9

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(B)

18

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(C)

18 \ln (2)

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(D) The limit doesn't exist

Factor out common factor: Rewrite the expression by factoring out the common factor $9\cdot2^1$ from the numerator. $\newline$ So, $\lim_{h \to 0}(9\cdot2^{1+h}-9\cdot2^{1})/h = \lim_{h \to 0}9\cdot2^1(2^h-1)/h$ .
Simplify expression: Simplify the expression by taking $9\times2^1$ out of the limit since it's a constant. $\newline$ So, $9\times2^1 \times \lim_{h \to 0}\frac{2^h-1}{h}.$
Calculate constant: Calculate $9 \times 2^1$ which is $9 \times 2 = 18$ . So, $18 \times \lim_{h \to 0}\frac{2^h-1}{h}$ .
Use derivative definition: Use the definition of the derivative for the function $f(x) = 2^x$ at $x = 1$ . The limit becomes $18 \cdot f'(1)$ where $f'(x) = \ln(2)\cdot2^x$ .
Calculate final result: Calculate $f'(1)$ which is $ext{ln}(2) imes 2^1 = ext{ln}(2) imes 2$ . So, $18 imes ext{ln}(2) imes 2$ .