# How do you find the limit of #lnx/x# as #x->oo#?

As we have:

The limit:

so:

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On request, giving a demonstration without using l'Hospital:

so:

based on the squeeze theorem.

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To find the limit of ln(x)/x as x approaches infinity, we can use L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating ln(x) with respect to x gives us 1/x, and differentiating x with respect to x gives us 1.

Now, we can evaluate the limit of the differentiated function:

lim(x->oo) (1/x) / 1

As x approaches infinity, 1/x approaches 0, and 1 remains as 1.

Therefore, the limit simplifies to:

lim(x->oo) 0 / 1

Which equals 0.

Hence, the limit of ln(x)/x as x approaches infinity is 0.

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To find the limit of ( \frac{\ln(x)}{x} ) as ( x ) approaches infinity, you can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator with respect to ( x ) separately, you get ( \lim_{x \to \infty} \frac{1/x}{1} ). This simplifies to ( \lim_{x \to \infty} \frac{1}{x} ), which approaches ( 0 ) as ( x ) goes to infinity. Therefore, the limit of ( \frac{\ln(x)}{x} ) as ( x ) approaches infinity is ( 0 ).

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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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