+26367 limit x to infinity 2x+3/5x+7
$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\
=\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\
=\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\
= \dfrac{2+0}{5+0} = \dfrac{2}{5}$$
+128408
If this is (2x+3)/(5x+7)...the limit of this as x approaches infinity is just (2/5)....remember..... we have a "same over the same" condition, here.....which means that, in a rational function, when the polynomials in the numerator and the denominator are of the same degree.....the function has a limit as x → ∞ = the ratio of the coefficients on the highest power of each polynomial.....
Here's a graph......
+26367 limit x to infinity 2x+3/5x+7
$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\
=\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\
=\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\
= \dfrac{2+0}{5+0} = \dfrac{2}{5}$$
