Question

1. in the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0/0. then, evaluate the limit.

(a) $lim_{x
ightarrow2}\frac{x^{2}+x - 6}{x^{2}-4}$
(b) $lim_{x
ightarrow - 3}\frac{sqrt{x + 7}-2}{x + 3}$
(c) $lim_{h
ightarrow0}\frac{(1 + h)^{3}-1}{h}$
(d) $lim_{x
ightarrow1}\frac{x^{3}-1}{x^{2}-1}$

Explanation:

Step1: Check indeterminate - form for (a)

Substitute \(x = 2\) into \(\frac{x^{2}+x - 6}{x^{2}-4}\).
When \(x = 2\), \(x^{2}+x - 6=2^{2}+2 - 6=4 + 2-6 = 0\) and \(x^{2}-4=2^{2}-4=0\).
Factor the numerator and denominator: \(x^{2}+x - 6=(x + 3)(x - 2)\) and \(x^{2}-4=(x + 2)(x - 2)\).
Then \(\lim_{x
ightarrow2}\frac{x^{2}+x - 6}{x^{2}-4}=\lim_{x
ightarrow2}\frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}=\lim_{x
ightarrow2}\frac{x + 3}{x + 2}\).
Substitute \(x = 2\) into \(\frac{x + 3}{x + 2}\), we get \(\frac{2+3}{2 + 2}=\frac{5}{4}\).

Step2: Check indeterminate - form for (b)

Substitute \(x=-3\) into \(\frac{\sqrt{x + 7}-2}{x + 3}\).
When \(x=-3\), \(\sqrt{x + 7}-2=\sqrt{-3 + 7}-2=\sqrt{4}-2=0\) and \(x + 3=-3 + 3=0\).
Rationalize the numerator: \(\frac{\sqrt{x + 7}-2}{x + 3}\times\frac{\sqrt{x + 7}+2}{\sqrt{x + 7}+2}=\frac{(x + 7)-4}{(x + 3)(\sqrt{x + 7}+2)}=\frac{x + 3}{(x + 3)(\sqrt{x + 7}+2)}=\frac{1}{\sqrt{x + 7}+2}\).
Substitute \(x=-3\) into \(\frac{1}{\sqrt{x + 7}+2}\), we get \(\frac{1}{\sqrt{-3 + 7}+2}=\frac{1}{2 + 2}=\frac{1}{4}\).

Step3: Check indeterminate - form for (c)

Substitute \(h = 0\) into \(\frac{(1 + h)^{3}-1}{h}\).
When \(h = 0\), \((1 + h)^{3}-1=(1+0)^{3}-1=0\) and \(h = 0\).
Expand \((1 + h)^{3}\) using the formula \((a + b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\), so \((1 + h)^{3}=1 + 3h+3h^{2}+h^{3}\).
Then \(\frac{(1 + h)^{3}-1}{h}=\frac{1 + 3h+3h^{2}+h^{3}-1}{h}=\frac{3h+3h^{2}+h^{3}}{h}=3 + 3h+h^{2}\).
Substitute \(h = 0\) into \(3 + 3h+h^{2}\), we get \(3\).

Step4: Check indeterminate - form for (d)

Substitute \(x = 1\) into \(\frac{x^{3}-1}{x^{2}-1}\).
When \(x = 1\), \(x^{3}-1=1^{3}-1=0\) and \(x^{2}-1=1^{2}-1=0\).
Factor \(x^{3}-1=(x - 1)(x^{2}+x + 1)\) and \(x^{2}-1=(x - 1)(x + 1)\).
Then \(\lim_{x
ightarrow1}\frac{x^{3}-1}{x^{2}-1}=\lim_{x
ightarrow1}\frac{(x - 1)(x^{2}+x + 1)}{(x - 1)(x + 1)}=\lim_{x
ightarrow1}\frac{x^{2}+x + 1}{x + 1}\).
Substitute \(x = 1\) into \(\frac{x^{2}+x + 1}{x + 1}\), we get \(\frac{1^{2}+1 + 1}{1+1}=\frac{3}{2}\).

Answer:

(a) \(\frac{5}{4}\)
(b) \(\frac{1}{4}\)
(c) \(3\)
(d) \(\frac{3}{2}\)