What is the limit as n approaches infinity of (ln(n+2))/(ln(n+1)) ?

The limit as n approaches infinity of (ln(n+2))/(ln(n+1)) is 1

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limit as n approaches infinity of (ln(n+2))/(ln(n+1))

n \to \infty lim (\frac{ln ( n + 2 )}{ln ( n + 1 )})

1

Solution steps

n \to \infty lim (\frac{ln ( n + 2 )}{ln ( n + 1 )})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{\frac{1}{n + 2}}{\frac{1}{n + 1}})

= n \to \infty lim (\frac{\frac{1}{n + 2}}{\frac{1}{n + 1}})

Simplify

\frac{\frac{1}{n + 2}}{\frac{1}{n + 1}} : \frac{n + 1}{n + 2}

= n \to \infty lim (\frac{n + 1}{n + 2})

Divide by highest denominator power:

\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}

= n \to \infty lim (\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}})

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

= \frac{lim _{n \to \infty} ( 1 + \frac{1}{n} )}{lim _{n \to \infty} ( 1 + \frac{2}{n} )}

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

n \to \infty lim (1 + \frac{2}{n}) = 1

= \frac{1}{1}

Simplify

= 1

x \to 1 lim (sin (- π x))

x \to 0 lim (\frac{1 - 2 sin ( x )}{tan ( x )})

x \to 3 lim (f (x) (3 - 2 x))

x \to 1 lim (\frac{x ^{22} - 2 x + 1}{x ^{3} - 2 x + 1})

x \to \infty lim ((+ x)^{2})

What is the limit as n approaches infinity of (ln(n+2))/(ln(n+1)) ?
The limit as n approaches infinity of (ln(n+2))/(ln(n+1)) is 1