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BlackZzzverrR [31]
2 years ago
6

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

(5x − ln(x))
Mathematics
1 answer:
Vesnalui [34]2 years ago
4 0

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 \lim_{x \to \infty} 5x - lnx

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

\lim_{x \to \ c} \frac{f_{x} }{g_{x} } =  \lim_{x \to \ c} \frac{f^{'}( x)}{g^{'} (x)}

\lim_{x \to \infty} 5x - lnx  can be written as

\lim_{x \to \infty} 5x - lnx\\=  \lim_{x \to \infty} x(5 - \frac{lnx}{x})

If we put the value of limit in lnx/x we get an indeterminate form ∞/∞.

Therefore, \lim_{x \to \infty}  \frac{lnx}{x}  = \frac{\frac{1}{x} }{1}

\implies \lim_{x \to \infty} \frac{1}{x} = 0   (as x tends to infinity 1/x tends to 0)

So,

\lim_{x \to \infty} 5x - lnx\\=  \lim_{x \to \infty} x(5 - \frac{lnx}{x})

=  \lim_{x \to \infty}x(5 -0)

=  \lim_{n \to \infty} 5x \\= \infty(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 ∞.

Find out more information about limit and L'hospital rule here:

brainly.com/question/14105620

#SPJ4

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