# How do you find the limit of #(1-[6/x])^x# as x approaches infinity using l'hospital's rule?

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To find the limit of (1-[6/x])^x as x approaches infinity using L'Hospital's rule, we can rewrite the expression as e^(x * ln(1-[6/x])). Taking the natural logarithm of (1-[6/x]) and applying L'Hospital's rule, we get the limit of ln(1-[6/x]) as x approaches infinity. Differentiating the numerator and denominator, we have (-6/x^2) / (-6/x^2). Simplifying this expression, we get 1. Therefore, the limit of ln(1-[6/x]) as x approaches infinity is 1. Finally, taking the exponential of both sides, we find that the limit of (1-[6/x])^x as x approaches infinity is e^1, which is equal to e.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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