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# limit x to infinity 2x+3/5x+7

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limit x to infinity 2x+3/5x+7

Guest Jan 27, 2015

#3
+19207
+5

limit x to infinity 2x+3/5x+7

$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\ =\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\ =\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\ = \dfrac{2+0}{5+0} = \dfrac{2}{5}$$

heureka  Jan 27, 2015
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#1
+92225
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could you rewrite this with brackets please?

Melody  Jan 27, 2015
#2
+85819
+5

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......

GRAPH

CPhill  Jan 27, 2015
#3
+19207
+5
$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\ =\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\ =\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\ = \dfrac{2+0}{5+0} = \dfrac{2}{5}$$