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

1) Compute the limit of

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

x \to +∞

The limit of

x^{4}

+∞

when

x \to +∞

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 +∞

The limit of

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

+∞

when

x \to +∞

The limit of

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

+∞

when

x \to +∞

For

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

, the limit takes the indeterminate form

\frac{+∞}{+∞}

when

x

→

+∞

2) To find the limit, we can factor out the dominant term and transform

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

into

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}

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

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

The inside of abs is non-negative: abs can be removed

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

To reduce the expression, we can replace the fraction:

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

with the new expression:

x

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

The limit of

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

0

when

x \to +∞

The limit of

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

1

when

x \to +∞

The limit of

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

1

when

x \to +∞

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 +∞

▷Result: $+∞$