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

Guest Jan 27, 2015

Best Answer 

 #3
avatar+18829 
+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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3+0 Answers

 #1
avatar+91477 
+5

could you rewrite this with brackets please?

Melody  Jan 27, 2015
 #2
avatar+81061 
+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
avatar+18829 
+5
Best Answer

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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