What is the limit as x approaches infinity of x(pi/4-arctan(x/(x+1))) ?

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limit as x approaches infinity of x(pi/4-arctan(x/(x+1)))

x \to \infty lim (x (\frac{π}{4} - arctan (\frac{x}{x + 1})))

\frac{1}{2}

Decimal

0.5

Solution steps

x \to \infty lim (x (\frac{π}{4} - arctan (\frac{x}{x + 1})))

Rewrite for L'Hopital

= x \to \infty lim (\frac{- arctan ( \frac{x}{1 + x} ) + \frac{π}{4}}{\frac{1}{x}})

Apply L'Hopital's Rule

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

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

Simplify

\frac{- \frac{1}{2 x ^{2} + 2 x + 1}}{- \frac{1}{x ^{2}}} : \frac{x ^{2}}{2 x ^{2} + 2 x + 1}

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

Divide by highest denominator power:

\frac{1}{2 + \frac{2}{x} + \frac{1}{x ^{2}}}

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

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 _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 2 + \frac{2}{x} + \frac{1}{x ^{2}} )}

x \to \infty lim (1) = 1

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

= \frac{1}{2}

x \to 0 + lim (2 cos (x))

x \to - 3 lim (\frac{4 x}{9 - x ^{2}})

x \to - 2 lim (\frac{8 + 4 x}{2 + x})

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

x \to 1 lim (3 x - 3)

What is the limit as x approaches infinity of x(pi/4-arctan(x/(x+1))) ?
The limit as x approaches infinity of x(pi/4-arctan(x/(x+1))) is 1/2