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Exercise 9.1 · Q4
Q.

Complete the table using a calculator and use the result to estimate the limit.

lim⁡x→−31−x−2x+3\lim_{x\to-3}\dfrac{\sqrt{1-x}-2}{x+3}

xx−3.1-3.1−3.01-3.01−3.0-3.0−2.999-2.999−2.99-2.99−2.9-2.9
f(x)f(x)
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Step 1. Try direct substitution. At x=−3x=-3: 1−(−3)−2=4−2=0\sqrt{1-(-3)}-2=\sqrt4-2=0 and x+3=0x+3=0, so 00\dfrac00, indeterminate.

Step 2. Substitute t=x+3t=x+3 to center the limit at 00. Then x=t−3x=t-3, so 1−x=1−(t−3)=4−t1-x=1-(t-3)=4-t, and

1−x−2x+3=4−t−2t.\dfrac{\sqrt{1-x}-2}{x+3}=\dfrac{\sqrt{4-t}-2}{t}.

Step 3. Rationalize by multiplying by the conjugate 4−t+2\sqrt{4-t}+2:

4−t−2t⋅4−t+24−t+2=(4−t)−4t(4−t+2)=−tt(4−t+2)=−14−t+2(t≠0).\dfrac{\sqrt{4-t}-2}{t}\cdot\dfrac{\sqrt{4-t}+2}{\sqrt{4-t}+2}=\dfrac{(4-t)-4}{t\left(\sqrt{4-t}+2\right)}=\dfrac{-t}{t\left(\sqrt{4-t}+2\right)}=\dfrac{-1}{\sqrt{4-t}+2}\quad(t\ne0).

Step 4. Table (using t=x+3t=x+3, so x=−3.1⇒t=−0.1x=-3.1\Rightarrow t=-0.1, etc.; the value x=−3x=-3 itself is excluded).

| xx | −3.1-3.1 | −3.01-3.01 | −2.999-2.999 | −2.99-2.99 | −2.9-2.9 | …

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