What is the limit as x approaches infinity of ln(x)+ln(1/(1/x+1)) ?

The limit as x approaches infinity of ln(x)+ln(1/(1/x+1)) is infinity

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limit as x approaches infinity of ln(x)+ln(1/(1/x+1))

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

\infty

Solution steps

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

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

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

x \to \infty lim (ln (x)) = \infty

x \to \infty lim (ln (\frac{1}{\frac{1}{x} + 1})) = 0

= \infty + 0

Apply Infinity Property:

\infty + c = \infty

= \infty

y^{2} (1 - x^{2})^{\frac{1}{2}} d y = arcsin (x) d x

g (u) = \frac{8 u ^{2}}{( u ^{2} + u ) ^{3}}

\int - x^{2} - 2 x + 3 d x

\int \frac{2 x + 1}{x ^{2} + x + 3} d x

f (x) = \frac{x ^{2} - 81}{x - 9}

What is the limit as x approaches infinity of ln(x)+ln(1/(1/x+1)) ?
The limit as x approaches infinity of ln(x)+ln(1/(1/x+1)) is infinity