What is the limit as x approaches infinity of sqrt(x+\sqrt{x)}-sqrt(x-\sqrt{x)} ?

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limit as x approaches infinity of sqrt(x+\sqrt{x)}-sqrt(x-\sqrt{x)}

x \to \infty lim (x + x - x - x)

1

Solution steps

x \to \infty lim (x + x - x - x)

Multiply by the conjugate of

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

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

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

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

Divide by

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

= 2 \cdot x \to \infty lim \frac{\frac{x}{x + 1}}{1 + \frac{x - x}{x + x}}

Simplify

\frac{x - x}{x + x} : \frac{x - x}{x + x}

= 2 \cdot x \to \infty lim \frac{\frac{x}{x + 1}}{1 + \frac{x - x}{x + 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

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

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

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

= 2 \cdot \frac{1}{2}

Simplify

2 \cdot \frac{1}{2} : 1

= 1

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

F (x) = 3 x^{4} - 65 x^{3}

t + 1

\int \frac{( 1 + cos ( 2 x ) )}{2} d x

y = \frac{2 x}{1 - cot ( x )}

What is the limit as x approaches infinity of sqrt(x+\sqrt{x)}-sqrt(x-\sqrt{x)} ?
The limit as x approaches infinity of sqrt(x+\sqrt{x)}-sqrt(x-\sqrt{x)} is 1