Evaluating limits of rational functions

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SUMMARY

The limit of the rational function (x^2 + 5) / (3x) as x approaches 0 is +∞, while the limit of x^2 / (3x) as x approaches 0 is 0. The presence of the constant +5 in the numerator causes the function to diverge to infinity because the numerator approaches 5 while the denominator approaches 0. This results in a very large quotient, confirming that the limit does not exist in the traditional sense, as infinity is not a real number. The analysis highlights the importance of understanding how constants affect limits in rational functions.

PREREQUISITES
  • Understanding of rational functions and limits
  • Familiarity with the concept of limits approaching infinity
  • Knowledge of polynomial behavior near specific points
  • Basic algebraic manipulation of fractions
NEXT STEPS
  • Study the concept of limits in calculus, focusing on limits approaching infinity
  • Learn about the behavior of rational functions near vertical asymptotes
  • Explore the significance of constants in polynomial limits
  • Investigate the application of L'Hôpital's Rule for indeterminate forms
USEFUL FOR

Students studying calculus, particularly those focusing on limits and rational functions, as well as educators looking for clear explanations of limit behavior in mathematical analysis.

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Homework Statement


Why does the limit as x approaches 0 of
x^2 + 5 / 3x go to infinity (with 0 as an essential disc.) but without the +5, the function goes to 0?


Homework Equations





The Attempt at a Solution


I tried approaching evaluating the limit of the function by comparing the exponents of the numerator and denominator and that seems to work if the +5 isn't there. What is the explanation for this difference?
 
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You mean (x^2+5)/(3x), right? Parentheses help. Without the +5, you've got x^2/(3x). That's simplifies to x/3, right? So it approaches 0. With the +5 the numerator approaches 5 and the denominator approaches 0, so the quotient is infinity. Counting powers doesn't really help here.
 
A little more precisely, for x very, very close to 0, the numerator is close to 5 and the denominator is close to 0. That will be a very, very, large number. As x gets closer to 0, the numerator stays close to 5 while the denominator gets closer to 0. That is, the limit is [itex]+\infty[/itex] (which, since [itex]\infty[/itex] is not a real number is the same as saying the limit does not exist.

(Comparing highest powers is useful when x goes to [itex]\pm\infty[/itex].)
 

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