Question

evaluate the given limits using the graph of the function ( f(x) = \frac{1}{(x + 2)^2} ) shown above. enter inf for ( infty ), -inf for ( -infty ), or dne if the limit does not exist, but is neither ( infty ) nor ( -infty ). if you are having a hard time seeing the picture clearly, click on the picture. it will expand to a larger picture on its own page so that you can inspect it more clearly. a) ( lim_{x \to -2} f(x) = square ) b) ( lim_{x \to -2^+} f(x) = square ) c) ( lim_{x \to -2} f(x) = square )

Explanation:

Part (a): $\lim_{x \to -2} f(x)$

Step1: Analyze the graph near \( x = -2 \)

The function \( f(x)=\frac{1}{(x + 2)^2} \) has a vertical asymptote at \( x=-2 \) (since the denominator is zero there and the numerator is non - zero). From the graph, as \( x \) approaches \( -2 \) from both the left and the right, the function values increase without bound.

Step2: Determine the limit

For a limit as \( x \to a \) to exist, the left - hand limit and the right - hand limit must be equal. Here, as \( x \) approaches \( -2 \) (both sides), \( f(x) \) goes to \( \infty \). So \( \lim_{x \to -2}f(x)=\text{INF} \)

Part (b): $\lim_{x \to -2^{-}} f(x)$ (Left - hand limit as \( x \) approaches \( -2 \))

Step1: Analyze the left - hand side of \( x=-2 \)

As \( x \) approaches \( -2 \) from the left (values of \( x \) less than \( -2 \), like \( x=-2 - h \) where \( h\to0^{+} \)), we substitute into \( f(x)=\frac{1}{(x + 2)^2} \). Let \( x=-2 - h \), then \( f(x)=\frac{1}{(-h)^2}=\frac{1}{h^2} \). As \( h\to0^{+} \), \( \frac{1}{h^2}\to\infty \). From the graph, as we move towards \( x = - 2 \) from the left, the function values shoot up to infinity.

Step2: Determine the left - hand limit

So \( \lim_{x \to -2^{-}}f(x)=\text{INF} \)

Part (c): $\lim_{x \to -2^{+}} f(x)$ (Right - hand limit as \( x \) approaches \( -2 \))

Step1: Analyze the right - hand side of \( x=-2 \)

As \( x \) approaches \( -2 \) from the right (values of \( x \) greater than \( -2 \), like \( x=-2 + h \) where \( h\to0^{+} \)), we substitute into \( f(x)=\frac{1}{(x + 2)^2} \). Let \( x=-2+h \), then \( f(x)=\frac{1}{h^2} \). As \( h\to0^{+} \), \( \frac{1}{h^2}\to\infty \). From the graph, as we move towards \( x=-2 \) from the right, the function values shoot up to infinity.

Step2: Determine the right - hand limit

So \( \lim_{x \to -2^{+}}f(x)=\text{INF} \)

Answer:

s:
a) \(\text{INF}\)

b) \(\text{INF}\)

c) \(\text{INF}\)