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

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limit as x approaches infinity of 1/(sqrt(x))-1/(sqrt(x+1))

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

0

Solution steps

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

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

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

= 0 - 0

Simplify

= 0

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

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

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

x \to - \frac{π}{2} - lim (- 6 x - \frac{π}{2})

x \to 0 + lim (x^{2} \cdot ln (x^{3}))

What is the limit as x approaches infinity of 1/(sqrt(x))-1/(sqrt(x+1)) ?
The limit as x approaches infinity of 1/(sqrt(x))-1/(sqrt(x+1)) is 0