Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→[infinity] ln(x) x

Respuesta :

The limit of [tex]\lim_{x \to \infty} \frac{lnx}{x}[/tex]  using L'Hospital rule is 0.

According to the given question.

We have to find the limit of ln(x)/x when x approaches to infinity.

If we let x = ∞, we get an indeterminate form [tex]\frac{\infty}{\infty}[/tex] ie. [tex]\frac{ln(x)}{x} = \frac{\infty}{\infty}[/tex].

And, we know that whenever we get indeterminate form ∞/∞ we apply L'Hospital rule.

Therefore, defferentiating numerator ln(x) and x with respect to x.

[tex]\implies \frac{d(ln(x))}{dx} = \frac{1}{x}[/tex] and [tex]\frac{d(x)}{dx} =1[/tex]

So,

[tex]\lim_{x\to \infty}\frac{\frac{1}{x} }{1}[/tex]

[tex]= \lim_{x \to \infty} \frac{1}{x}[/tex]

As, x tends to ∞, 1/x tends to 0. Because ∞ is very large number and 1 divided by a very large number always approaches to 0 and it is very close to zero.

Therefore,

[tex]\lim_{x \to \infty} \frac{1}{x} = 0[/tex]

Hence, the limit of [tex]\lim_{x \to \infty} \frac{lnx}{x}[/tex] is 0.

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