Estimating the limit of an oscillating function near a point

As $x \to 0$, the quantity $1/x$ grows without bound, so it sweeps through all real numbers infinitely many times, causing $\cos(1/x)$ to cycle repeatedly between $-1$ and $1$. For example, choosing $x = \frac{1}{2n\pi}$ gives $f(x) = \cos(2n\pi) = 1$, while choosing $x = \frac{1}{(2n+1)\pi}$ gives $f(x) = \cos((2n+1)\pi) = -1$. Both of these sequences of $x$-values approach $0$, yet the corresponding $f(x)$ values approach two different numbers.

Since we can find $x$-values arbitrarily close to $0$ where $f(x)$ is as close to $1$ as we like, and other $x$-values equally close to $0$ where $f(x)$ is as close to $-1$ as we like, $f(x)$ never settles down to a single value $L$ near $x=0$. This matches the graph, which shows increasingly rapid oscillation as $x \to 0$. Therefore the limit does not exist due to oscillation.

It oscillates. As x approaches zero, cosine of one over x repeatedly cycles between negative one and one without settling on a single value. This is different from a function growing without bound, since the values here stay confined between negative one and one at all times. The graph in part a shows increasingly rapid back-and-forth oscillation near zero, confirming this is the cause of nonexistence.