Question

evaluate $lim_{x
ightarrow0}sinleft(\frac{1}{4x^{2}}
ight)$ using the following table of values. enter an exact answer. if the limit does not exist, submit your answer as $varnothing$. provide your answer below: $lim_{x
ightarrow0}sinleft(\frac{1}{4x^{2}}
ight)=square$

Explanation:

Step1: Analyze the limit behavior

As $x\to0$, the argument $\frac{1}{4x^{2}}\to+\infty$. The sine - function $y = \sin(t)$ oscillates between $- 1$ and $1$ as $t\to+\infty$. That is, as $x$ approaches $0$, $\sin(\frac{1}{4x^{2}})$ does not approach a single finite value. So the limit $\lim_{x\to0}\sin(\frac{1}{4x^{2}})$ does not exist.

Answer:

$\varnothing$