Question

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))

Answer 1

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:

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