1. Let's start by understanding the **definition of a limit**: A limit $\,\lim_{x \to a} f(x) = L$ exists if as $x$ approaches a value $a$, the function $f(x)$ approaches a single finite number $L$.\n\n2. Why a limit might **not** exist? There are three main reasons:\n - The left-hand limit and right-hand limit are different.\n - The function grows without bound (tends to infinity or negative infinity).\n - The function oscillates endlessly near $a$ without settling on a value.\n\n3. **Example 1: Limit exists**\nConsider $f(x) = 2x + 3$, find $\lim_{x \to 1} f(x)$.\nStep 1: Plug $x=1$ directly or evaluate the limit since $f$ is continuous: $f(1) = 2(1) + 3 = 5$.\nAnswer: $\lim_{x \to 1} (2x+3) = 5$. The limit exists and equals 5 because both sides approach 5.\n\n4. **Example 2: Limit does not exist due to different left and right limits**\nConsider the piecewise function:\n$$f(x) = \begin{cases} 1, & x<0 \\ 2, & x \geq 0 \end{cases}$$\nFind $\lim_{x \to 0} f(x)$.\nStep 1: Compute left-hand limit: $\lim_{x \to 0^-} f(x) = 1$.\nStep 2: Compute right-hand limit: $\lim_{x \to 0^+} f(x) = 2$.\nStep 3: Since left and right limits are different, $\lim_{x \to 0} f(x)$ **does not exist**.\n\n5. **Example 3: Limit does not exist due to infinite growth**\nConsider $f(x) = \frac{1}{x^2}$, find $\lim_{x \to 0} f(x)$.\nStep 1: As $x$ approaches 0, $\frac{1}{x^2}$ grows without bound towards $+\infty$.\nAnswer: The limit does not exist because the function tends to infinity (unbounded).\n\n6. **Example 4: Limit does not exist due to oscillation**\nConsider $f(x) = \sin \frac{1}{x}$, find $\lim_{x \to 0} f(x)$.\nStep 1: As $x$ approaches 0, $\frac{1}{x}$ grows very large, and $\sin \frac{1}{x}$ oscillates rapidly between -1 and 1.\nAnswer: The limit does not exist because $f(x)$ does not approach any single value.