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let f(x) = (x^2 - 4)/(x + 2) (a) calculate f(x) for each value of x in …

Question

let f(x) = (x^2 - 4)/(x + 2)
(a) calculate f(x) for each value of x in the following table
(type integers or decimals rounded to four decimal places as needed)

x(x^2 - 4)/(x + 2)x(x^2 - 4)/(x + 2)
-1.99-2.01
-1.999-2.001
-1.9999-2.0001

(b) make a conjecture about the value of lim(x→ - 2) (x^2 - 4)/(x + 2)

Explanation:

Step1: Simplify the function

Given \(f(x)=\frac{x + 2}{x^{2}-4}\), factor the denominator \(x^{2}-4=(x + 2)(x - 2)\). So \(f(x)=\frac{x + 2}{(x + 2)(x - 2)}=\frac{1}{x - 2}\), for \(x
eq - 2\).

Step2: Calculate \(f(x)\) for different \(x\) - values

When \(x=-1.9\):

Substitute \(x=-1.9\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-1.9)=\frac{1}{-1.9-2}=\frac{1}{-3.9}\approx - 0.2564\).

When \(x=-1.99\):

Substitute \(x=-1.99\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-1.99)=\frac{1}{-1.99-2}=\frac{1}{-3.99}\approx - 0.2506\).

When \(x=-1.999\):

Substitute \(x=-1.999\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-1.999)=\frac{1}{-1.999-2}=\frac{1}{-3.999}\approx - 0.2500\).

When \(x=-1.9999\):

Substitute \(x=-1.9999\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-1.9999)=\frac{1}{-1.9999-2}=\frac{1}{-3.9999}\approx - 0.2500\).

When \(x=-2.0001\):

Substitute \(x=-2.0001\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-2.0001)=\frac{1}{-2.0001-2}=\frac{1}{-4.0001}\approx - 0.2500\).

When \(x=-2.001\):

Substitute \(x=-2.001\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-2.001)=\frac{1}{-2.001-2}=\frac{1}{-4.001}\approx - 0.2499\).

When \(x=-2.01\):

Substitute \(x=-2.01\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-2.01)=\frac{1}{-2.01-2}=\frac{1}{-4.01}\approx - 0.2494\).

When \(x=-2.1\):

Substitute \(x=-2.1\) into \(f(x)=\frac{1}{x - 2}\), we get \(f(-2.1)=\frac{1}{-2.1-2}=\frac{1}{-4.1}\approx - 0.2439\).

Step3: Make a conjecture about the limit

As \(x\) approaches \(-2\) from both the left - hand side (\(x\lt - 2\)) and the right - hand side (\(x\gt - 2\)), the values of \(f(x)\) approach \(-0.25\). So \(\lim_{x
ightarrow - 2}\frac{x + 2}{x^{2}-4}=-\frac{1}{4}=-0.25\).

Answer:

\(x\)\(f(x)\)
\(-1.99\)\(-0.2506\)
\(-1.999\)\(-0.2500\)
\(-1.9999\)\(-0.2500\)
\(-2.0001\)\(-0.2500\)
\(-2.001\)\(-0.2499\)
\(-2.01\)\(-0.2494\)
\(-2.1\)\(-0.2439\)

\(\lim_{x
ightarrow - 2}\frac{x + 2}{x^{2}-4}=-0.25\)