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I'm trying to resolve this limit without using De L'Hopital rule: $$ \lim_{x\to 0}\frac{2^{x+1}-2}{x} $$ With the rule i get ln(4).

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Well, you can note that $$2^{x+1}-2=2\cdot2^x-2=2\left(2^x-1\right)=2\cdot\left(2^x-2^0\right).$$ Hence: $$\lim_{x\to 0}\frac{2^{x+1}-2}x=2\cdot\lim_{x\to0}\frac{2^x-2^0}{x-0}=2\cdot\frac{d}{dx}\left[2^x\right]_{x=0}$$

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This is the derivative of $2^{x+1}$ at $x=0$.

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$$2=e^{\ln2}\implies 2^x=e^{x\ln2}$$ and $$\lim_{h\to0}\frac{e^h-1}h=1$$

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$$\begin{array}{l} f\left( x \right) = 2^{x + 1} = e^{\left( {x + 1} \right)\ln 2} \\ f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{e^{\left( {x + 1} \right)\ln 2} - 2}}{{x - 0}} = 2\ln 2 \\ \end{array}$$

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