lim(sqrt(16*x^4+sin(x))/x, x->+inf)

1) Let's evaluate the limit of

\frac{\sqrt{16 x^{4} + \sin (x)}}{x}

when

x \to +∞

ⓘ A quantity that grows without bound raised to a positive power still grows without bound, to a negative power it tends to 0

The limit of

x^{4}

+∞

when

x \to +∞

ⓘ A quantity that grows without bound multiplied by a quantity with a nonzero limit still grows without bound, the signs giving +∞ or -∞

The limit of

16 x^{4}

+∞

when

x \to +∞

The limit of

\sin (x)

does not exist but is bounded in

[-1, 1]

when

x \to +∞

ⓘ A quantity that grows without bound plus a finite or bounded quantity still grows without bound

The limit of

16 x^{4} + \sin (x)

+∞

when

x \to +∞

ⓘ We use a reference limit of the function sqrt

The limit of

\sqrt{16 x^{4} + \sin (x)}

+∞

when

x \to +∞

For

\frac{\sqrt{16 x^{4} + \sin (x)}}{x}

, we obtain the indeterminate form

\frac{+∞}{+∞}

x

tends to

+∞

2) To determine the limit, factor out the dominant term to rewrite

16 x^{4} + \sin (x)

16 x^{4} \cdot (1 + \frac{\sin (x)}{16 x^{4}})

\frac{\sqrt{16 x^{4} \cdot (1 + \frac{\sin (x)}{16 x^{4}})}}{x}

ⓘ The root of a product is the product of the roots,

\sqrt{a b} = \sqrt{a} \sqrt{b}

\sqrt{16 x^{4} \cdot (1 + \frac{\sin (x)}{16 x^{4}})}

\sqrt{16} \cdot \sqrt{x^{4}} \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}

\frac{\sqrt{16} \cdot \sqrt{x^{4}} \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}}{x}

ⓘ The square root of a nonnegative real number is the nonnegative number whose square is that number:

\sqrt{16}

4

since

4^{2}

16

■Taking the square root of a square gives the absolute value:

\sqrt{x^{4}}

| x^{2} |

\frac{4 \cdot | x^{2} | \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}}{x}

For a nonnegative expression, the absolute value equals the expression

ⅰ

| x^{2} |

x^{2}

\frac{4 x^{2} \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}}{x}

The fraction:

\frac{x^{2}}{x}

simplifies by the common factor:

x

into the new expression:

x

4 \cdot x \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}

ⓘ A quantity that grows without bound multiplied by a quantity with a nonzero limit still grows without bound, the signs giving +∞ or -∞■A bounded quantity divided by a quantity that grows without bound tends to 0

The limit of

\frac{\sin (x)}{16 x^{4}}

0

when

x \to +∞

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

1 + \frac{\sin (x)}{16 x^{4}}

1

when

x \to +∞

ⓘ The function sqrt is continuous, its limit is its value at the limit of its argument

The limit of

\sqrt{1 + \frac{\sin (x)}{16 x^{4}}}

\sqrt{1}

when

x \to +∞

Let's calculate:

\sqrt{1}

1

ⓘ A quantity that grows without bound multiplied by a quantity with a nonzero limit still grows without bound, the signs giving +∞ or -∞

The limit of

4 x \cdot \sqrt{1 + \frac{\sin (x)}{16 x^{4}}}

+∞

when

x \to +∞

So the limit of

\frac{\sqrt{16 x^{4} + \sin (x)}}{x}

+∞

when

x \to +∞

▶Limit:

+∞