lim_(x rarr0)(1-2x)^((1)/(x))

Recognize Indeterminate Form: To solve the limit lim_(x -> 0)(1-2x)^(1/x), we can recognize that this is an indeterminate form of type 0^0. To resolve this, we can use the property that e^(lim_(x -> 0)(f(x) * ln(g(x)))) = lim_(x -> 0)(g(x)^(f(x))) if lim_(x -> 0)(f(x) * ln(g(x))) exists. We will rewrite the expression in a form that allows us to apply this property.Rewrite Using Natural Logarithm: Let's rewrite the limit using the natural logarithm: lim_(x -> 0)(1-2x)^(1/x) = e^(lim_(x -> 0)(1/x * ln(1-2x))). Now we need to find the limit inside the exponent: lim_(x -> 0)(1/x * ln(1-2x)).Apply L'Hôpital's Rule: We can now apply L'Hôpital's Rule to the limit inside the exponent because it is an indeterminate form of type 0/0. To do this, we differentiate the numerator and the denominator separately: Let's find the derivative of the numerator, ln(1-2x), with respect to x: d/dx[ln(1-2x)] = -2/(1-2x). And the derivative of the denominator, 1/x, with respect to x: d/dx[1/x] = -1/x^2.Evaluate Limit: Now we apply L'Hôpital's Rule: lim_(x -> 0)(1/x * ln(1-2x)) = lim_(x -> 0)(-2/(1-2x) * (-x^2)) = lim_(x -> 0)(2x^2/(1-2x)). We can now evaluate this limit as x approaches 0.Simplify Expression: Plugging in x = 0 into the simplified expression 2x^2/(1-2x), we get: 2*0^2/(1-2*0) = 0/1 = 0. So, lim_(x -> 0)(1/x * ln(1-2x)) = 0.Final Answer: Now that we have the limit of the exponent, we can go back to the original limit: lim_(x -> 0)(1-2x)^(1/x) = e^(lim_(x -> 0)(1/x * ln(1-2x))) = e^0.We know that e^0 = 1. Therefore, the final answer is: lim_(x -> 0)(1-2x)^(1/x) = 1.

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Algebra 2

Sum of finite series not start from 1

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\lim_{x \to 0}(1-2x)^{\frac{1}{x}}

Recognize Indeterminate Form: To solve the limit $\lim_{x \to 0}(1-2x)^{\frac{1}{x}}$ , we can recognize that this is an indeterminate form of type $0^0$ . To resolve this, we can use the property that $e^{\lim_{x \to 0}(f(x) \cdot \ln(g(x)))} = \lim_{x \to 0}(g(x)^{f(x)})$ if $\lim_{x \to 0}(f(x) \cdot \ln(g(x)))$ exists. We will rewrite the expression in a form that allows us to apply this property.
Rewrite Using Natural Logarithm: Let's rewrite the limit using the natural logarithm: $\newline$ $\lim_{x \to 0}(1-2x)^{\frac{1}{x}} = e^{\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x))}$ . $\newline$ Now we need to find the limit inside the exponent: $\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x))$ .
Apply L'Hôpital's Rule: We can now apply L'Hôpital's Rule to the limit inside the exponent because it is an indeterminate form of type $0/0$ . To do this, we differentiate the numerator and the denominator separately: $\newline$ Let's find the derivative of the numerator, $\ln(1-2x)$ , with respect to $x$ : $\frac{d}{dx}[\ln(1-2x)] = -\frac{2}{1-2x}$ . $\newline$ And the derivative of the denominator, $1/x$ , with respect to $x$ : $\frac{d}{dx}[1/x] = -\frac{1}{x^2}$ .
Evaluate Limit: Now we apply L'Hôpital's Rule: $\newline$ $\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x)) = \lim_{x \to 0}(\frac{-2}{1-2x} \cdot (-x^2)) = \lim_{x \to 0}(\frac{2x^2}{1-2x})$ . $\newline$ We can now evaluate this limit as $x$ approaches $0$ .
Simplify Expression: Plugging in $x = 0$ into the simplified expression $\frac{2x^2}{1-2x}$ , we get: $\frac{2\cdot 0^2}{1-2\cdot 0} = \frac{0}{1} = 0$ . So, $\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x)) = 0$ .
Final Answer: Now that we have the limit of the exponent, we can go back to the original limit: $\lim_{x \to 0}(1-2x)^{\frac{1}{x}} = e^{\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x))} = e^0$ .
Final Answer: Now that we have the limit of the exponent, we can go back to the original limit: $\newline$ $\lim_{x \to 0}(1-2x)^{\frac{1}{x}} = e^{\lim_{x \to 0}(\frac{1}{x} \cdot \ln(1-2x))} = e^0$ .We know that $e^0 = 1$ . Therefore, the final answer is: $\newline$ $\lim_{x \to 0}(1-2x)^{\frac{1}{x}} = 1$ .