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Sagot :
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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