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For the function
f(x) given below, evaluate
lim_(x rarr oo)f(x) and
lim_(x rarr-oo)f(x).

f(x)=5sin(4x^(3))

For the function $f(x)$ given below, evaluate $\lim _{x \rightarrow \infty} f(x)$ and $\lim _{x \rightarrow-\infty} f(x)$ . $\newline$ $f(x)=5 \sin \left(4 x^{3}\right)$

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Find derivatives of sine and cosine functions

Full solution

Q. For the function

f(x)

given below, evaluate

\lim _{x \rightarrow \infty} f(x)

and

\lim _{x \rightarrow-\infty} f(x)

.

\newline

f(x)=5 \sin \left(4 x^{3}\right)

Limit as $x$ approaches infinity: First, let's consider the limit as $x$ approaches infinity. $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}5\sin(4x^3)$ Since the sine function oscillates between $-1$ and $1$ , the value of $\sin(4x^3)$ will also oscillate between $-1$ and $1$ as $x$ approaches infinity.
Oscillation with scaling: The coefficient $5$ does not affect the oscillation, it just scales the amplitude of the sine function. $\newline$ So, the limit does not exist because the function keeps oscillating and does not settle to a single value.
Limit as $x$ approaches negative infinity: Now, let's consider the limit as $x$ approaches negative infinity. $\newline$ $\lim_{x \to -\infty}f(x) = \lim_{x \to -\infty}5\sin(4x^3)$ $\newline$ The reasoning is the same as for the limit as $x$ approaches infinity. $\newline$ The sine function will still oscillate between $-1$ and $1$ , regardless of whether $x$ is going to positive or negative infinity.

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