Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 6: limit laws, the squeeze theorem, limits at infinity, the epsilon-delta definition, continuity and the IVT第 1 至 6 节：极限法则、夹逼定理、无穷极限、epsilon-delta 定义、连续性与介值定理CALC I
Evaluate (a) $\lim_{x\to 3}\frac{x^{2}-x-6}{x-3}$; (b) $\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}$; (c) $\lim_{x\to 2}\frac{|x-2|}{x-2}$.计算 (a) $\lim_{x\to 3}\frac{x^{2}-x-6}{x-3}$；(b) $\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}$；(c) $\lim_{x\to 2}\frac{|x-2|}{x-2}$。
Direct substitution gives $\tfrac{0}{0}$, so factor. The numerator factors as $x^{2}-x-6=(x-3)(x+2)$. (M1)直接代入得 $\tfrac{0}{0}$ 不定式，故进行因式分解。分子 $x^{2}-x-6=(x-3)(x+2)$。(M1)
$$ \lim_{x\to 3}\frac{(x-3)(x+2)}{x-3}=\lim_{x\to 3}(x+2)=5. $$(A1)
Multiply numerator and denominator by $\sqrt{x+4}+2$: (M1)分子分母同乘 $\sqrt{x+4}+2$：(M1)
$$ \frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\frac{(x+4)-4}{x\left(\sqrt{x+4}+2\right)}=\frac{1}{\sqrt{x+4}+2}. $$Hence the limit is $\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{4}$. (A1)故极限为 $\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{4}$。(A1)
For $x>2$, $|x-2|=x-2$, so the quotient is $+1$; for $x<2$, $|x-2|=-(x-2)$, so the quotient is $-1$. (M1)当 $x>2$ 时，$|x-2|=x-2$，商为 $+1$；当 $x<2$ 时，$|x-2|=-(x-2)$，商为 $-1$。(M1)
The right-hand limit is $1$ and the left-hand limit is $-1$. Since they differ, the two-sided limit does not exist. (A1)右侧极限为 $1$，左侧极限为 $-1$。两者不相等，故双侧极限不存在。(A1)
Evaluate (a) $\lim_{x\to\infty}\frac{3x^{2}-5x+1}{2x^{2}+7}$; (b) $\lim_{x\to\infty}\frac{\sqrt{4x^{2}+x}}{x+1}$; (c) $\lim_{x\to-\infty}\frac{2x}{\sqrt{x^{2}+1}}$.计算 (a) $\lim_{x\to\infty}\frac{3x^{2}-5x+1}{2x^{2}+7}$；(b) $\lim_{x\to\infty}\frac{\sqrt{4x^{2}+x}}{x+1}$；(c) $\lim_{x\to-\infty}\frac{2x}{\sqrt{x^{2}+1}}$。
Divide top and bottom by $x^{2}$: $\dfrac{3-5/x+1/x^{2}}{2+7/x^{2}}$. (M1) As $x\to\infty$ the $1/x$ terms vanish, leaving $\dfrac{3}{2}$. (A1)分子分母同除以 $x^{2}$，得 $\dfrac{3-5/x+1/x^{2}}{2+7/x^{2}}$。(M1) 当 $x\to\infty$ 时，含 $1/x$ 的项趋于零，结果为 $\dfrac{3}{2}$。(A1)
For $x\to\infty$, $\sqrt{4x^{2}+x}=x\sqrt{4+1/x}$ since $x>0$. Dividing by $x$: (M1)当 $x\to\infty$ 时，因 $x>0$，有 $\sqrt{4x^{2}+x}=x\sqrt{4+1/x}$。除以 $x$：(M1)
$$ \frac{\sqrt{4+1/x}}{1+1/x}\longrightarrow \frac{\sqrt{4}}{1}=2. $$(A1)
Here $\sqrt{x^{2}+1}=|x|\sqrt{1+1/x^{2}}$, and for $x\to-\infty$ we have $|x|=-x$. (M1)此处 $\sqrt{x^{2}+1}=|x|\sqrt{1+1/x^{2}}$，当 $x\to-\infty$ 时 $|x|=-x$。(M1)
$$ \frac{2x}{-x\sqrt{1+1/x^{2}}}=\frac{2}{-\sqrt{1+1/x^{2}}}\longrightarrow \frac{2}{-1}=-2. $$(A1)
Evaluate (a) $\lim_{x\to 0}\frac{\sin 5x}{3x}$; (b) $\lim_{x\to 0} x^{2}\cos(1/x)$ by the squeeze theorem; (c) $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}$.计算 (a) $\lim_{x\to 0}\frac{\sin 5x}{3x}$；(b) 用夹逼定理求 $\lim_{x\to 0} x^{2}\cos(1/x)$；(c) $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}$。
Write the quotient so the argument of $\sin$ matches the denominator inside it: (M1)改写商式，使 $\sin$ 的参数与分母内的因子一致：(M1)
$$ \frac{\sin 5x}{3x}=\frac{5}{3}\cdot\frac{\sin 5x}{5x}\longrightarrow \frac{5}{3}\cdot 1=\frac{5}{3}. $$(A1)
For all $x\ne 0$, $-1\le\cos(1/x)\le 1$, so multiplying by $x^{2}\ge 0$: (M1)对所有 $x\ne 0$，$-1\le\cos(1/x)\le 1$，乘以 $x^{2}\ge 0$：(M1)
$$ -x^{2}\le x^{2}\cos(1/x)\le x^{2}. $$(M1) Both $-x^{2}\to 0$ and $x^{2}\to 0$ as $x\to 0$, so by the squeeze theorem the middle expression also tends to $0$. (A1)(M1) 当 $x\to 0$ 时，$-x^{2}\to 0$ 且 $x^{2}\to 0$，由夹逼定理，中间表达式也趋于 $0$。(A1)
Multiply by $\dfrac{1+\cos x}{1+\cos x}$: (M1)乘以 $\dfrac{1+\cos x}{1+\cos x}$：(M1)
$$ \frac{1-\cos x}{x^{2}}=\frac{1-\cos^{2}x}{x^{2}(1+\cos x)}=\frac{\sin^{2}x}{x^{2}(1+\cos x)}=\left(\frac{\sin x}{x}\right)^{2}\frac{1}{1+\cos x}. $$As $x\to 0$ this is $1^{2}\cdot\dfrac{1}{1+1}=\dfrac{1}{2}$. (A1)当 $x\to 0$ 时，结果为 $1^{2}\cdot\dfrac{1}{1+1}=\dfrac{1}{2}$。(A1)
For $f(x)=\frac{x^{2}-4}{x-2}$ ($x<2$), $a$ ($x=2$), $bx+1$ ($x>2$): (a) the left limit; (b) values of $a,b$ for continuity at $2$; (c) classify the discontinuity of $\frac{x^{2}-4}{x-2}$.对于 $f(x)=\frac{x^{2}-4}{x-2}$（$x<2$），$a$（$x=2$），$bx+1$（$x>2$）：(a) 求左极限；(b) 求使 $f$ 在 $x=2$ 处连续的 $a,b$ 值；(c) 分类 $\frac{x^{2}-4}{x-2}$ 的间断类型。
For $x<2$, $\dfrac{x^{2}-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$. (M1) So $\lim_{x\to 2^{-}}f(x)=2+2=4$. (A1)当 $x<2$ 时，$\dfrac{x^{2}-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$。(M1) 故 $\lim_{x\to 2^{-}}f(x)=2+2=4$。(A1)
Continuity at $2$ requires $\lim_{x\to 2^{-}}f=\lim_{x\to 2^{+}}f=f(2)$. (M1) The left limit is $4$ and $f(2)=a$, so $a=4$. (A1)$f$ 在 $x=2$ 处连续要求 $\lim_{x\to 2^{-}}f=\lim_{x\to 2^{+}}f=f(2)$。(M1) 左极限为 $4$，$f(2)=a$，故 $a=4$。(A1)
The right limit is $\lim_{x\to 2^{+}}(bx+1)=2b+1$. (M1) Setting $2b+1=4$ gives $b=\tfrac{3}{2}$. (A1)右极限为 $\lim_{x\to 2^{+}}(bx+1)=2b+1$。(M1) 令 $2b+1=4$，得 $b=\tfrac{3}{2}$。(A1)
The bare expression $\dfrac{x^{2}-4}{x-2}$ is undefined at $x=2$ but its two-sided limit exists and equals $4$. (A1)表达式 $\dfrac{x^{2}-4}{x-2}$ 在 $x=2$ 处无定义，但其双侧极限存在且等于 $4$。(A1)
Because the limit exists finitely, the gap is a single missing point: a removable discontinuity, repaired by defining the value to be $4$. (R1)因为极限有限存在，缺陷只是单个缺失点，即可去间断点，通过定义函数值为 $4$ 即可修复。(R1)
Prove by $\varepsilon$-$\delta$: (a) $\lim_{x\to 4}(2x+3)=11$; (b) $\lim_{x\to 3}x^{2}=9$.用 $\varepsilon$-$\delta$ 方法证明：(a) $\lim_{x\to 4}(2x+3)=11$；(b) $\lim_{x\to 3}x^{2}=9$。
Let $\varepsilon>0$. We analyse the target inequality to find $\delta$: $|(2x+3)-11|=|2x-8|=2|x-4|$. (M1)设 $\varepsilon>0$。分析目标不等式以确定 $\delta$：$|(2x+3)-11|=|2x-8|=2|x-4|$。(M1)
This is less than $\varepsilon$ exactly when $|x-4|<\tfrac{\varepsilon}{2}$, so choose $\delta=\tfrac{\varepsilon}{2}$. (A1)当且仅当 $|x-4|<\tfrac{\varepsilon}{2}$ 时上式小于 $\varepsilon$，故取 $\delta=\tfrac{\varepsilon}{2}$。(A1)
Verification: if $0<|x-4|<\delta$ then $|(2x+3)-11|=2|x-4|<2\delta=\varepsilon$. Hence the limit is $11$. (R1)验证：若 $0<|x-4|<\delta$，则 $|(2x+3)-11|=2|x-4|<2\delta=\varepsilon$。故极限为 $11$。(R1)
Let $\varepsilon>0$. Then $|x^{2}-9|=|x-3|\,|x+3|$. The factor $|x+3|$ is not constant, so we first cap it by restricting $\delta\le 1$. (M1)设 $\varepsilon>0$。则 $|x^{2}-9|=|x-3|\,|x+3|$。因子 $|x+3|$ 非常数，故先限制 $\delta\le 1$ 以控制其大小。(M1)
If $|x-3|<1$ then $2
To force $7|x-3|<\varepsilon$ we need $|x-3|<\tfrac{\varepsilon}{7}$. Choose $\delta=\min\!\left(1,\tfrac{\varepsilon}{7}\right)$. (A1)为使 $7|x-3|<\varepsilon$，需 $|x-3|<\tfrac{\varepsilon}{7}$。取 $\delta=\min\!\left(1,\tfrac{\varepsilon}{7}\right)$。(A1)
Verification: if $0<|x-3|<\delta$ then both $|x+3|<7$ (from $\delta\le 1$) and $|x-3|<\tfrac{\varepsilon}{7}$, so $|x^{2}-9|<7\cdot\tfrac{\varepsilon}{7}=\varepsilon$. (R1)验证：若 $0<|x-3|<\delta$，则 $|x+3|<7$（由 $\delta\le 1$）且 $|x-3|<\tfrac{\varepsilon}{7}$，故 $|x^{2}-9|<7\cdot\tfrac{\varepsilon}{7}=\varepsilon$。(R1)
(a) $x^{3}-4x+1=0$ has a root in $(0,1)$; (b) $\cos x=x$ has a solution in $(0,\tfrac{\pi}{2})$; (c) every continuous $f:[0,1]\to[0,1]$ has a fixed point.(a) $x^{3}-4x+1=0$ 在 $(0,1)$ 内有实根；(b) $\cos x=x$ 在 $(0,\tfrac{\pi}{2})$ 内有解；(c) 每个连续映射 $f:[0,1]\to[0,1]$ 都有不动点。
Let $f(x)=x^{3}-4x+1$. It is a polynomial, hence continuous on $[0,1]$. (M1) Evaluate the endpoints: $f(0)=1>0$ and $f(1)=1-4+1=-2<0$. (A1)令 $f(x)=x^{3}-4x+1$。$f$ 为多项式，故在 $[0,1]$ 上连续。(M1) 计算端点值：$f(0)=1>0$，$f(1)=1-4+1=-2<0$。(A1)
Since $f$ is continuous and changes sign, the IVT guarantees a $c\in(0,1)$ with $f(c)=0$. (R1)因 $f$ 连续且变号，由介值定理保证存在 $c\in(0,1)$ 使得 $f(c)=0$。(R1)
Let $g(x)=\cos x-x$, continuous everywhere. (M1) Then $g(0)=\cos 0-0=1>0$ and $g\!\left(\tfrac{\pi}{2}\right)=\cos\tfrac{\pi}{2}-\tfrac{\pi}{2}=0-\tfrac{\pi}{2}=-\tfrac{\pi}{2}<0$. (A1)令 $g(x)=\cos x-x$，处处连续。(M1) 则 $g(0)=\cos 0-0=1>0$，$g\!\left(\tfrac{\pi}{2}\right)=\cos\tfrac{\pi}{2}-\tfrac{\pi}{2}=0-\tfrac{\pi}{2}=-\tfrac{\pi}{2}<0$。(A1)
By the IVT there is a $c\in\left(0,\tfrac{\pi}{2}\right)$ with $g(c)=0$, i.e. $\cos c=c$. (R1)由介值定理，存在 $c\in\left(0,\tfrac{\pi}{2}\right)$ 使得 $g(c)=0$，即 $\cos c=c$。(R1)
Define $h(x)=f(x)-x$ on $[0,1]$; it is continuous as a difference of continuous functions. (M1)定义 $h(x)=f(x)-x$ 在 $[0,1]$ 上，作为连续函数之差，$h$ 连续。(M1)
Because $f$ maps into $[0,1]$: $h(0)=f(0)-0=f(0)\ge 0$ and $h(1)=f(1)-1\le 0$. (M1)因 $f$ 映射至 $[0,1]$：$h(0)=f(0)-0=f(0)\ge 0$，$h(1)=f(1)-1\le 0$。(M1)
If $h(0)=0$ then $c=0$ is a fixed point; if $h(1)=0$ then $c=1$ is. Otherwise $h(0)>0$ and $h(1)<0$, so the IVT gives $c\in(0,1)$ with $h(c)=0$. (A1) In every case $f(c)=c$. (R1)若 $h(0)=0$，则 $c=0$ 为不动点；若 $h(1)=0$，则 $c=1$ 为不动点。否则 $h(0)>0$ 且 $h(1)<0$，由介值定理存在 $c\in(0,1)$ 使 $h(c)=0$。(A1) 三种情形下均有 $f(c)=c$。(R1)
From $\cos x\le\frac{\sin x}{x}\le 1$ on $0<|x|<\tfrac{\pi}{2}$: (a) prove $\lim_{x\to 0}\frac{\sin x}{x}=1$; (b) hence evaluate $\lim_{x\to 0}\frac{1-\cos x}{x}$.由 $0<|x|<\tfrac{\pi}{2}$ 时的 $\cos x\le\frac{\sin x}{x}\le 1$：(a) 证明 $\lim_{x\to 0}\frac{\sin x}{x}=1$；(b) 由此计算 $\lim_{x\to 0}\frac{1-\cos x}{x}$。
The given inequality $\cos x\le\dfrac{\sin x}{x}\le 1$ holds for $0<|x|<\tfrac{\pi}{2}$. (M1)已知不等式 $\cos x\le\dfrac{\sin x}{x}\le 1$ 在 $0<|x|<\tfrac{\pi}{2}$ 时成立。(M1)
As $x\to 0$, the lower bound $\cos x\to\cos 0=1$ and the upper bound is the constant $1$. (A1)当 $x\to 0$ 时，下界 $\cos x\to\cos 0=1$，上界为常数 $1$。(A1)
Both bounds tend to the same value $1$, so by the squeeze theorem the trapped quantity satisfies $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$. (R1) The two-sided conclusion is valid because $\tfrac{\sin x}{x}$ is even, so the left and right behaviour match. (A1)两个界趋向同一值 $1$，由夹逼定理，被夹量满足 $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$。(R1) 双侧结论成立，因为 $\tfrac{\sin x}{x}$ 是偶函数，左右行为一致。(A1)
Multiply by $\dfrac{1+\cos x}{1+\cos x}$: (M1)乘以 $\dfrac{1+\cos x}{1+\cos x}$：(M1)
$$ \frac{1-\cos x}{x}=\frac{1-\cos^{2}x}{x(1+\cos x)}=\frac{\sin^{2}x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}. $$(M1) As $x\to 0$, the first factor tends to $1$ by part (a), while the second tends to $\dfrac{0}{1+1}=0$. (A1) The product tends to $1\cdot 0=0$.(M1) 当 $x\to 0$ 时，第一因子由 (a) 趋于 $1$，第二因子趋于 $\dfrac{0}{1+1}=0$。(A1) 乘积趋于 $1\cdot 0=0$。
For $f(x)=\frac{2x^{2}-2}{x^{2}-x-2}$: (a) factor and state the domain; (b) the removable discontinuity and its hole; (c) the vertical asymptote and one-sided behaviour; (d) the horizontal asymptote.对于 $f(x)=\frac{2x^{2}-2}{x^{2}-x-2}$：(a) 因式分解并写出定义域；(b) 可去间断点及空洞坐标；(c) 铅直渐近线及单侧行为；(d) 水平渐近线。
$2x^{2}-2=2(x-1)(x+1)$ and $x^{2}-x-2=(x-2)(x+1)$. (M1) The denominator is zero at $x=2$ and $x=-1$, so the domain is all reals except $x=2$ and $x=-1$. (A1)$2x^{2}-2=2(x-1)(x+1)$，$x^{2}-x-2=(x-2)(x+1)$。(M1) 分母在 $x=2$ 和 $x=-1$ 处为零，故定义域为除 $x=2$ 和 $x=-1$ 外的所有实数。(A1)
The factor $(x+1)$ cancels, so for $x\ne -1$, $f(x)=\dfrac{2(x-1)}{x-2}$. (M1) Because the factor cancels, the discontinuity at $x=-1$ is removable. The hole sits at the limiting height (M1)因子 $(x+1)$ 可以消去，故当 $x\ne -1$ 时 $f(x)=\dfrac{2(x-1)}{x-2}$。(M1) 因因子消去，$x=-1$ 处的间断为可去间断点。空洞位于极限高度处：(M1)
$$ \lim_{x\to -1}\frac{2(x-1)}{x-2}=\frac{2(-2)}{-3}=\frac{-4}{-3}=\frac{4}{3}. $$So there is a hole at $\left(-1,\tfrac{4}{3}\right)$. (A1)故空洞在 $\left(-1,\tfrac{4}{3}\right)$。(A1)
The factor $(x-2)$ does not cancel, so $x=2$ is a vertical asymptote. Use the reduced form $\dfrac{2(x-1)}{x-2}$, whose numerator at $x=2$ is $2(1)=2>0$. (M1)因子 $(x-2)$ 不消去，故 $x=2$ 为铅直渐近线。用约简后的形式 $\dfrac{2(x-1)}{x-2}$，其分子在 $x=2$ 处为 $2(1)=2>0$。(M1)
As $x\to 2^{+}$, the denominator $\to 0^{+}$, so $f\to+\infty$. (A1) As $x\to 2^{-}$, the denominator $\to 0^{-}$, so $f\to-\infty$. (A1)当 $x\to 2^{+}$ 时，分母 $\to 0^{+}$，故 $f\to+\infty$。(A1) 当 $x\to 2^{-}$ 时，分母 $\to 0^{-}$，故 $f\to-\infty$。(A1)
Degrees of numerator and denominator are equal, so the limit at infinity is the ratio of leading coefficients: (M1)分子与分母次数相同，故无穷处的极限为最高次系数之比：(M1)
$$ \lim_{x\to\pm\infty}\frac{2x^{2}-2}{x^{2}-x-2}=\frac{2}{1}=2. $$The horizontal asymptote is $y=2$. (A1)水平渐近线为 $y=2$。(A1)
For $f(x)=\sqrt{x}$: (a) show $f'(a)=\frac{1}{2\sqrt{a}}$ from the limit definition; (b) the tangent line at $a=9$; (c) confirm $f'(9)$ with the $x\to a$ form.对于 $f(x)=\sqrt{x}$：(a) 由极限定义证明 $f'(a)=\frac{1}{2\sqrt{a}}$；(b) 在 $a=9$ 处的切线方程；(c) 用 $x\to a$ 形式确认 $f'(9)$。
$f'(a)=\displaystyle\lim_{h\to 0}\frac{\sqrt{a+h}-\sqrt{a}}{h}$. Multiply by the conjugate $\dfrac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}$: (M1)$f'(a)=\displaystyle\lim_{h\to 0}\frac{\sqrt{a+h}-\sqrt{a}}{h}$。乘以共轭 $\dfrac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}$：(M1)
$$ \frac{(a+h)-a}{h\left(\sqrt{a+h}+\sqrt{a}\right)}=\frac{h}{h\left(\sqrt{a+h}+\sqrt{a}\right)}=\frac{1}{\sqrt{a+h}+\sqrt{a}}. $$(M1·A1 for the cancellation of $h$) Letting $h\to 0$ gives $\dfrac{1}{\sqrt{a}+\sqrt{a}}=\dfrac{1}{2\sqrt{a}}$. (A1)（$h$ 消去得 M1·A1）令 $h\to 0$ 得 $\dfrac{1}{\sqrt{a}+\sqrt{a}}=\dfrac{1}{2\sqrt{a}}$。(A1)
$f(9)=3$ and $f'(9)=\dfrac{1}{2\sqrt{9}}=\dfrac{1}{6}$. (M1) Point-slope form: $y-3=\tfrac{1}{6}(x-9)$. (A1)$f(9)=3$，$f'(9)=\dfrac{1}{2\sqrt{9}}=\dfrac{1}{6}$。(M1) 点斜式：$y-3=\tfrac{1}{6}(x-9)$。(A1)
Simplify: $y=\tfrac{1}{6}x-\tfrac{9}{6}+3=\tfrac{1}{6}x-\tfrac{3}{2}+3=\tfrac{1}{6}x+\tfrac{3}{2}$. (A1)化简：$y=\tfrac{1}{6}x-\tfrac{9}{6}+3=\tfrac{1}{6}x-\tfrac{3}{2}+3=\tfrac{1}{6}x+\tfrac{3}{2}$。(A1)
$f'(9)=\displaystyle\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}$. Factor the denominator as a difference of squares: $x-9=(\sqrt{x}-3)(\sqrt{x}+3)$. (M1)$f'(9)=\displaystyle\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}$。将分母因式分解为平方差：$x-9=(\sqrt{x}-3)(\sqrt{x}+3)$。(M1)
$$ \frac{\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{1}{\sqrt{x}+3}\longrightarrow\frac{1}{3+3}=\frac{1}{6}. $$(M1·A1) This matches part (b). (A1)（M1·A1）与 (b) 一致。(A1)
For $g(x)=\frac{\sqrt{x+c}-3}{x-4}$: (a) find $c$ giving a finite limit at $x=4$; (b) evaluate it; (c) the value $g(4)$ for continuity.对于 $g(x)=\frac{\sqrt{x+c}-3}{x-4}$：(a) 求使 $x=4$ 处极限有限的 $c$；(b) 计算该极限；(c) 求使 $g$ 在 $x=4$ 连续的 $g(4)$ 值。
As $x\to 4$ the denominator $x-4\to 0$. For a finite limit the numerator must also tend to $0$ (otherwise the quotient is a nonzero number over zero, which diverges). (M1)当 $x\to 4$ 时，分母 $x-4\to 0$。为使极限有限，分子也必须趋于 $0$（否则商为非零数除以零，发散）。(M1)
Require $\sqrt{4+c}-3=0$, i.e. $\sqrt{4+c}=3$, so $4+c=9$ and $c=5$. (A1) Any other $c$ leaves a nonzero numerator at $x=4$, forcing an infinite (or non-existent) limit. (R1)令 $\sqrt{4+c}-3=0$，即 $\sqrt{4+c}=3$，故 $4+c=9$，$c=5$。(A1) 其他 $c$ 值均使分子在 $x=4$ 处非零，导致极限无穷大（或不存在）。(R1)
With $c=5$, multiply by $\dfrac{\sqrt{x+5}+3}{\sqrt{x+5}+3}$: (M1)取 $c=5$，乘以 $\dfrac{\sqrt{x+5}+3}{\sqrt{x+5}+3}$：(M1)
$$ \frac{\sqrt{x+5}-3}{x-4}=\frac{(x+5)-9}{(x-4)\left(\sqrt{x+5}+3\right)}=\frac{x-4}{(x-4)\left(\sqrt{x+5}+3\right)}=\frac{1}{\sqrt{x+5}+3}. $$(M1) As $x\to 4$ this is $\dfrac{1}{\sqrt{9}+3}=\dfrac{1}{6}$. (A1)(M1) 当 $x\to 4$ 时，结果为 $\dfrac{1}{\sqrt{9}+3}=\dfrac{1}{6}$。(A1)
A function is continuous at a point when its value equals its limit there, so set $g(4)=\dfrac{1}{6}$. (A1)函数在某点连续当且仅当函数值等于该点处的极限，故令 $g(4)=\dfrac{1}{6}$。(A1)