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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

Find $\lim _{h \rightarrow 0} \frac{9 \cdot 2^{1+h}-9 \cdot 2^{1}}{h}$ $\newline$ Choose $1$ answer: $\newline$ (A) $9$ $\newline$ (B) $18$ $\newline$ (C) $18 \ln (2)$ $\newline$ (D) The limit doesn't exist

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Multiply and divide rational expressions

Full solution

Q. Find

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

\newline

Choose

1

answer:

\newline

(A)

9

\newline

(B)

18

\newline

(C)

18 \ln (2)

\newline

(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, $18 imes ext{ln}(2) imes 2$ .

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