Question

evaluate the given limits using the graph of the function ( f(x) = \frac{3}{e^x + 1} ) 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 -infty} f(x) = square ) b) ( lim_{x \to +infty} f(x) = square ) c) ( lim_{x \to 0^-} f(x) = square ) d) ( lim_{x \to 0^+} f(x) = square ) e) ( lim_{x \to 0} f(x) = square )

Explanation:

Part (a)

Step1: Analyze \( e^x \) as \( x \to -\infty \)

As \( x \to -\infty \), \( e^x \to 0 \) (since exponential function with base \( e>1 \) approaches 0 as \( x \to -\infty \)).

Step2: Substitute into \( f(x) \)

\( f(x)=\frac{3}{e^x + 1} \), so when \( e^x \to 0 \), \( f(x)\to\frac{3}{0 + 1}=3 \).

Step1: Analyze \( e^x \) as \( x \to +\infty \)

As \( x \to +\infty \), \( e^x \to +\infty \) (exponential growth).

Step2: Substitute into \( f(x) \)

\( f(x)=\frac{3}{e^x + 1} \), so \( e^x + 1\to +\infty \), and \( \frac{3}{+\infty}\to 0 \).

Step1: Analyze left - hand limit as \( x \to 0^- \)

For \( x \to 0^- \), we can also use the function formula \( f(x)=\frac{3}{e^x + 1} \). Substitute \( x = 0 \) (since the function is continuous at \( x = 0 \), left - hand limit equals the function value at \( x = 0 \)). \( e^0=1 \), so \( f(0)=\frac{3}{1 + 1}=\frac{3}{2}=1.5 \). From the graph, as \( x \) approaches 0 from the left, the function approaches \( \frac{3}{2} \).

Step2: Conclusion

\( \lim_{x\to 0^-}f(x)=\frac{3}{e^0+1}=\frac{3}{2} = 1.5 \)

Answer:

\( 3 \)

Part (b)