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

The limit as x approaches infinity of x(arctan(x)-pi/2) is -1

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limit as x approaches infinity of x(arctan(x)-pi/2)

x \to \infty lim (x (arctan (x) - \frac{π}{2}))

- 1

Solution steps

x \to \infty lim (x (arctan (x) - \frac{π}{2}))

Rewrite for L'Hopital

= x \to \infty lim (\frac{arctan ( x ) - \frac{π}{2}}{\frac{1}{x}})

Apply L'Hopital's Rule

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

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

Simplify

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

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

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

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

Divide by highest denominator power:

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

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

x \to \infty lim (1) = 1

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

= - \frac{1}{1}

Simplify

= - 1

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