Step by step solution to "limx→∞(xsin2πx2+1)"

brovaddallaunnei-4603

Answered question

2022-07-11

$\lim_{x \to \infty} (x \sin (2 π \sqrt{x^{2} + 1}))$

nick1337

Expert2023-05-28Added 777 answers

To evaluate the limit

{lim}_{x \to \infty} (x \sin (2 π \sqrt{x^{2} + 1}))

, we can use some properties of limits and trigonometric functions.
As

x

approaches infinity, the term

\sqrt{x^{2} + 1}

can be approximated as

x

since the

x^{2}

term dominates as

x

grows large.
Therefore, we can rewrite the expression as:

{lim}_{x \to \infty} (x \sin (2 π \sqrt{x^{2} + 1})) = {lim}_{x \to \infty} (x \sin (2 π x))

Since

\sin (2 π x)

is a periodic function with a period of

2 π

, the value of

\sin (2 π x)

oscillates between -1 and 1 as

x

increases without bound.
As a result, the expression

x \sin (2 π x)

does not converge to a specific value as

x

approaches infinity. Instead, it oscillates between negative and positive infinity.
Therefore, the limit is not defined.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?