Question

example: ( f(x)=+5-\frac{x}{x^{2}-3} ) (( \frac{x}{x^{2}-3} approx 0 ) for large ( x ))
end behavior is ( y = 5 )
example: ( g(x)=x + 1+\frac{5}{x} ) (( \frac{5}{x} approx 0 ) for large ( x ))
end behavior is ( y = x + 1 )
example: ( h(x)=-4 x^{2}+\frac{x - 3}{x^{2}-5} ) (( \frac{x - 3}{x^{2}-5} approx 0 ) for large ( x ))
end behavior is ( y=-4 x^{2} )
your turn:
( f(x)=2 x+\frac{-11}{5 x} ) end behavior: ( y= )
( g(x)=-5 x^{2}+1+\frac{1}{x^{3}} ) end behavior: ( y= )
( h(x)=\frac{-10}{x + 1} ) end behavior: ( y= )
if nothing here, then what value can be put here?
( k(x)=\frac{5}{9}+\frac{2}{9 x} ) end behavior: ( y= )

Explanation:

Step1: Analyze the end - behavior of \( f(x)=2x+\frac{-11}{5x} \)

For large \( x \), the term \(\frac{-11}{5x}\approx0\). So the end - behavior is dominated by the \( 2x \) term.

Step2: Analyze the end - behavior of \( g(x)=-5x^{2}+1+\frac{1}{x^{3}} \)

For large \( x \), the terms \(1\) and \(\frac{1}{x^{3}}\approx0\). So the end - behavior is dominated by the \(-5x^{2}\) term.

Step3: Analyze the end - behavior of \( h(x)=\frac{-10}{x + 1} \)

For large \( x \), \(h(x)=\frac{-10}{x+1}\approx\frac{-10}{x}\approx0\) (using the fact that for \(x\to\pm\infty\), \(x + 1\approx x\)).

Step4: Analyze the end - behavior of \(k(x)=\frac{5}{9}+\frac{2}{9x}\)

For large \( x \), the term \(\frac{2}{9x}\approx0\). So the end - behavior is dominated by the \(\frac{5}{9}\) term.

Answer:

1. \( y = 2x \)
2. \( y=-5x^{2} \)
3. \( y = 0 \)
4. \( y=\frac{5}{9} \)