What is the limit as x approaches infinity of x(sqrt(x^2+4)-x) ?

The limit as x approaches infinity of x(sqrt(x^2+4)-x) is 2

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limit as x approaches infinity of x(sqrt(x^2+4)-x)

x \to \infty lim (x (x^{2} + 4 - x))

2

Solution steps

x \to \infty lim (x (x^{2} + 4 - x))

Multiply by the conjugate of

x^{2} + 4 - x : \frac{4}{x ^{2} + 4 + x}

= x \to \infty lim (x \frac{4}{x ^{2} + 4 + x})

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

= 4 \cdot x \to \infty lim (x \frac{1}{x ^{2} + 4 + x})

Simplify

x \frac{1}{x ^{2} + 4 + x} : \frac{x}{x ^{2} + 4 + x}

= 4 \cdot x \to \infty lim (\frac{x}{x ^{2} + 4 + x})

Divide by highest denominator power:

\frac{1}{1 + \frac{4}{x ^{2}} + 1}

= 4 \cdot x \to \infty lim \frac{1}{1 + \frac{4}{x ^{2}} + 1}

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

= 4 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 1 + \frac{4}{x ^{2}} + 1 )}

x \to \infty lim (1) = 1

x \to \infty lim (1 + \frac{4}{x ^{2}} + 1) = 2

= 4 \cdot \frac{1}{2}

Simplify

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

= 2

y = 5 x e^{- k x}

x y^{^{'}} + (1 + x) = e^{- x}

\int sin (2 x) - 5 cos (4 x) d x

\int 10 x^{5} e^{x^{6}} d x

\frac{\partial}{\partial x} (ln (e^{x} + e^{- x}))

What is the limit as x approaches infinity of x(sqrt(x^2+4)-x) ?
The limit as x approaches infinity of x(sqrt(x^2+4)-x) is 2