This problem tests defining limits and using limit notation for functions with known limit properties like sin(x)/x. The expression \(\lim_{x \to 0} u(x) = 1\) correctly states the well-known limit of sin(x)/x approaching 1 as x nears 0. Despite u(0)=3, the notation focuses on values around x=0, not at it. This exemplifies standard limits overriding point definitions. A tempting distractor like \(\lim_{x \to 0} u(x) = 3\) fails because it confuses the redefined function value with the actual limit. For limit notation, recall standard limits, specify approach to the point, equate to the known value, and separate from f(a).

This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to1} h(x)=1$, is valid because the limit of $\ln(x)/(x-1)$ as $x$ approaches 1 is 1, a standard form. It's $0/0$ indeterminate, resolved by L'Hôpital to $1/x$ over 1, approaching 1. Both sides agree. A tempting distractor is choice A, $\lim_{x\to1} h(x)=0$, from numerator but ignoring derivative. Use $\to$ and apply rules for indeterminates.

This question tests limit notation for trigonometric limits. The limit as x approaches 0 of s(x) is 3 because the expression simplifies to 3 times sin(3x)/(3x), which approaches 3*1=3 using the standard limit of sin(u)/u as u approaches 0. The function value s(0)=0 does not affect the limit, which examines values near 0. Therefore, the notation \(\lim_{x\to0} s(x)=3\) is correct. A tempting distractor is A, which mistakenly uses the function value at x=0 for the limit. A checklist for limit notation includes specifying the function, the variable approaching a value, checking one-sided limits agree, and stating the limit value or that it does not exist.

This question tests the skill of defining limits and using limit notation. The expression $\lim_{x\to0}$ n(x)=\frac{1}{2}$ is correct because rationalizing yields ($\sqrt{x+1}$-1)/x = 1/($\sqrt{x+1}$+1) → 1/2. Both sides approach 1/2. This notation equals the limit value. A tempting distractor is C, n(0)=\frac{1}{2}$, but n(0) undefined. To check limit notation, rationalize expressions, use →, and compute the value.

This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to2} f(x)$ does not exist, is valid because the left limit is 6 and the right is 5, differing. Left uses $3x$ approaching 6, right $x^2+1$ approaching 5. Mismatch means no two-sided limit. A tempting distractor is choice A, $\lim_{x\to2} f(x)=7$, confusing with the point value. Use $\to$ and evaluate one-sided for piecewise.

This question tests the skill of defining limits and using limit notation. The expression $\lim_{x\to1}$ q(x)=2 is correct because simplifying to x+1 for x≠1 approaches 2. The redefinition at 1=0 doesn't affect limit. This notation denotes the approached value. A tempting distractor is B, q(1)=0, but that's the value, not limit. To check limit notation, factor rationals, use →, and ignore point value.

This question tests the skill of defining limits and using limit notation. The expression in choice B, $\lim_{x\to3} f(x)=6$, is valid because simplifying the function gives $f(x) = x+3$ for $x\ne3$, which approaches 6 as $x$ approaches 3 from both sides. The limit focuses on the behavior near $x=3$, not the value at $x=3$, which is defined as 1. Even though there is a hole at $x=3$, the function values get arbitrarily close to 6 as $x$ gets close to 3. A tempting distractor is choice A, $\lim_{x\to3} f(x)=1$, which confuses the limit with the function value at the point. Always check the limit by simplifying expressions and evaluating the approach, and use the $\to$ symbol in the subscript for proper notation.

This question tests the skill of defining limits and using limit notation. The expression $\lim_{x\to0}$ p(x)=4 correctly represents the limit because from the left, x+4 approaches 4, and from the right, 4-x approaches 4. Both pieces agree at 4 as x nears 0, even though the function is defined differently on each side. This notation indicates the common value approached. A tempting distractor is B, p(0)=0, but p(0)=4-0=4, and it confuses function value with limit. To check limit notation, ensure the approach is denoted by →, the limit value is consistent from both sides, and proper syntax is used.

This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to0} n(x)$ does not exist, is valid because as $x$ approaches 0 from the left, $n(x)$ goes to $-infty$, and from the right to $+infty$. The one-sided limits do not agree on a finite value. The behavior diverges in opposite directions. A tempting distractor is choice A, $\lim_{x\to0} n(x)=0$, perhaps thinking of the pole but ignoring infinity. Use $\to$ and state 'does not exist' for divergent cases.

This problem tests defining limits and using limit notation for algebraic functions with potential discontinuities. The expression $\lim_{x \to 3} h(x) = 6$ correctly describes the behavior as x approaches 3, since simplifying $\frac{x^2-9}{x-3}$ yields $x+3$, which nears 6. Even though $h(3)=10$ is defined differently, the limit notation captures the nearby values, not the point itself. This illustrates removable discontinuities where limits exist despite redefinition. A tempting choice like $\lim_{x \to 3} h(x) = 10$ fails because it incorrectly uses the function value at x=3 instead of the approached value. Remember in limit notation to specify the variable and approach point, equate to the value from surrounding points, simplify expressions if needed, and separate from f(a).