What is the limit as x approaches infinity of x-sqrt(x^2-16) ?

The limit as x approaches infinity of x-sqrt(x^2-16) is 0

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limit as x approaches infinity of x-sqrt(x^2-16)

x \to \infty lim (x - x^{2} - 16)

0

Solution steps

x \to \infty lim (x - x^{2} - 16)

Multiply by the conjugate of

x - x^{2} - 16 : \frac{16}{x + x ^{2} - 16}

= x \to \infty lim (\frac{16}{x + x ^{2} - 16})

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

= 16 \cdot x \to \infty lim (\frac{1}{x + x ^{2} - 16})

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

= 16 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( x + x ^{2} - 16 )}

x \to \infty lim (1) = 1

x \to \infty lim (x + x^{2} - 16) = \infty

= 16 \cdot \frac{1}{\infty}

Simplify

16 \cdot \frac{1}{\infty} : 0

= 0

x \to 0 + lim (∣ x ∣^{x})

x \to \infty lim (\frac{20}{x})

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

x \to 1 - lim (2 x)

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

What is the limit as x approaches infinity of x-sqrt(x^2-16) ?
The limit as x approaches infinity of x-sqrt(x^2-16) is 0