What is the limit as x approaches infinity of 10x(ln(x+8)-ln(x)) ?

The limit as x approaches infinity of 10x(ln(x+8)-ln(x)) is 80

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limit as x approaches infinity of 10x(ln(x+8)-ln(x))

x \to \infty lim (10 x (ln (x + 8) - ln (x)))

80

Solution steps

x \to \infty lim (10 x (ln (x + 8) - ln (x)))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 10 \cdot x \to \infty lim (x (ln (x + 8) - ln (x)))

Rewrite for L'Hopital

= 10 \cdot x \to \infty lim (\frac{ln ( 8 + x ) - ln ( x )}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{- \frac{8}{x ( x + 8 )}}{- \frac{1}{x ^{2}}})

= 10 \cdot x \to \infty lim (\frac{- \frac{8}{x ( x + 8 )}}{- \frac{1}{x ^{2}}})

Simplify

\frac{- \frac{8}{x ( x + 8 )}}{- \frac{1}{x ^{2}}} : \frac{8 x}{x + 8}

= 10 \cdot x \to \infty lim (\frac{8 x}{x + 8})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 10 \cdot 8 \cdot x \to \infty lim (\frac{x}{x + 8})

Divide by highest denominator power:

\frac{1}{1 + \frac{8}{x}}

= 10 \cdot 8 \cdot x \to \infty lim (\frac{1}{1 + \frac{8}{x}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= 10 \cdot 8 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 1 + \frac{8}{x} )}

x \to \infty lim (1) = 1

x \to \infty lim (1 + \frac{8}{x}) = 1

= 10 \cdot 8 \cdot \frac{1}{1}

Simplify

= 80

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What is the limit as x approaches infinity of 10x(ln(x+8)-ln(x)) ?
The limit as x approaches infinity of 10x(ln(x+8)-ln(x)) is 80