calculate the following limits by direct substitution.
$lim_{x
ightarrow0}sqrt{3(x^{2}+12)}=square$
$lim_{a
ightarrow - 10}\frac{(a + 7)^{4}}{a + 1}=square$
$lim_{a
ightarrow - 4}\frac{a^{2}-3a + 4}{a - 12}=square$
$lim_{s
ightarrow8}sqrt{\frac{13 - s}{s + 12}}=square$
$lim_{t
ightarrow5}\frac{(1 - t)(t + 5)}{3t - 7}=square$
$lim_{b
ightarrow7}\frac{24}{b + 1}-(b - 4)^{2}=square$

Question

Accepted Answer

# Explanation:
## Step1: Substitute $x = 0$ into $\sqrt{3(x^{2}+12)}$
$$
\begin{align*}
\lim_{x
ightarrow0}\sqrt{3(x^{2}+12)}&=\sqrt{3(0^{2}+12)}\
&=\sqrt{3\times12}\
&=\sqrt{36}\
& = 6
\end{align*}
$$
## Step2: Substitute $a=- 10$ into $\frac{(a + 7)^{4}}{a + 1}$
$$
\begin{align*}
\lim_{a
ightarrow - 10}\frac{(a + 7)^{4}}{a + 1}&=\frac{(-10 + 7)^{4}}{-10+1}\
&=\frac{(-3)^{4}}{-9}\
&=\frac{81}{-9}\
&=-9
\end{align*}
$$
## Step3: Substitute $a=-4$ into $\frac{a^{2}-3a + 4}{a - 12}$
$$
\begin{align*}
\lim_{a
ightarrow - 4}\frac{a^{2}-3a + 4}{a - 12}&=\frac{(-4)^{2}-3\times(-4)+4}{-4-12}\
&=\frac{16 + 12+4}{-16}\
&=\frac{32}{-16}\
&=-2
\end{align*}
$$
## Step4: Substitute $s = 8$ into $\sqrt{\frac{13 - s}{s + 12}}$
$$
\begin{align*}
\lim_{s
ightarrow8}\sqrt{\frac{13 - s}{s + 12}}&=\sqrt{\frac{13-8}{8 + 12}}\
&=\sqrt{\frac{5}{20}}\
&=\sqrt{\frac{1}{4}}\
&=\frac{1}{2}
\end{align*}
$$
## Step5: Substitute $t = 5$ into $\frac{(1 - t)(t + 5)}{3t-7}$
$$
\begin{align*}
\lim_{t
ightarrow5}\frac{(1 - t)(t + 5)}{3t-7}&=\frac{(1-5)(5 + 5)}{3\times5-7}\
&=\frac{(-4)\times10}{15 - 7}\
&=\frac{-40}{8}\
&=-5
\end{align*}
$$
## Step6: Substitute $b = 7$ into $\frac{24}{b + 1}-(b - 4)^{2}$
$$
\begin{align*}
\lim_{b
ightarrow7}\frac{24}{b + 1}-(b - 4)^{2}&=\frac{24}{7+1}-(7 - 4)^{2}\
&=\frac{24}{8}-3^{2}\
&=3 - 9\
&=-6
\end{align*}
$$
# Answer:
1. $6$
2. $-9$
3. $-2$
4. $\frac{1}{2}$
5. $-5$
6. $-6$

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QUESTION IMAGE

Question

Explanation:

Step1: Substitute \(x = 0\) into \(\sqrt{3(x^{2}+12)}\)

Step2: Substitute \(a=- 10\) into \(\frac{(a + 7)^{4}}{a + 1}\)

Step3: Substitute \(a=-4\) into \(\frac{a^{2}-3a + 4}{a - 12}\)

Step4: Substitute \(s = 8\) into \(\sqrt{\frac{13 - s}{s + 12}}\)

Step5: Substitute \(t = 5\) into \(\frac{(1 - t)(t + 5)}{3t-7}\)

Step6: Substitute \(b = 7\) into \(\frac{24}{b + 1}-(b - 4)^{2}\)

Answer: