Question

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

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. \(-6\)