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Sagot :
The limit of lim x→[infinity] (5x − ln(x)) by using L'hospital rule is ∞.
According to the given question.
We have to find the limit of [tex]\lim_{x \to \infty} 5x - lnx[/tex]
As we know that L'hospital rule is a theorem which provides a technique to evaluate limits of indeterminate forms.
And the formual for L'hospital rule is
[tex]\lim_{x \to \ c} \frac{f_{x} }{g_{x} } = \lim_{x \to \ c} \frac{f^{'}( x)}{g^{'} (x)}[/tex]
[tex]\lim_{x \to \infty} 5x - lnx[/tex] can be written as
[tex]\lim_{x \to \infty} 5x - lnx\\= \lim_{x \to \infty} x(5 - \frac{lnx}{x})[/tex]
If we put the value of limit in lnx/x we get an indeterminate form ∞/∞.
Therefore, [tex]\lim_{x \to \infty} \frac{lnx}{x} = \frac{\frac{1}{x} }{1}[/tex]
[tex]\implies \lim_{x \to \infty} \frac{1}{x} = 0[/tex] (as x tends to infinity 1/x tends to 0)
So,
[tex]\lim_{x \to \infty} 5x - lnx\\= \lim_{x \to \infty} x(5 - \frac{lnx}{x})[/tex]
[tex]= \lim_{x \to \infty}x(5 -0)[/tex]
[tex]= \lim_{n \to \infty} 5x \\= \infty[/tex](as x tends to ∞ 5x also tends to infinity)
Therefore, the limit of lim x→[infinity] (5x − ln(x)) by using L'hospital rule is ∞.
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