1. HaǐTH lim _(xarrow -infty )(1+(2)/(x))^x+5

Question

Question

1. HaǐTH lim _(xarrow -infty )(1+(2)/(x))^x+5

Answer 1

\(\lim_{x \to -\infty}(1+\frac{2}{x})^x + 5 = \frac{1}{e^2} + 5\)

Answer 2

The given problem is to find the limit as \(x\) approaches negative infinity of the expression \((1+\frac{2}{x})^x + 5\). This is a classic limit problem involving an exponential expression. To solve this, we need to recognize that as \(x\) approaches negative infinity, \(\frac{2}{x}\) approaches 0. Therefore, the base \(1 + \frac{2}{x}\) of the exponential expression approaches 1. The expression \((1 + \frac{2}{x})^x\) can be compared to the well-known limit \(\lim_{x \to \infty}(1 + \frac{1}{x})^x\), which is the definition of \(e\), the base of natural logarithms. However, in our case, since \(x\) approaches negative infinity, the expression \((1 + \frac{2}{x})^x\) approaches \(1/e^2\) because the exponent is negative, which flips the base of the exponential expression. Finally, we add 5 to this limit, as indicated in the original expression, to get the final result.

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1. HaǐTH lim _(xarrow -infty )(1+(2)/(x))^x+5

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