Question

How do I determine the end behaviour of the graph, \[f\left( x \right)=\dfrac{\left( 3x-3 \right)}{\left( 4x+5 \right)}\] in limit notation?

Answer 1

Hint: Problems like these are very easy to solve once we know the concepts, equations and other formulae in depth and detail. We must also recall the limit from first principle as it is one of the core things for limits, discontinuity and functions. In such problems we check for limits as ‘x’ approaches infinity, if it tends to some value then there is a horizontal asymptote. An asymptote is one which meets the curve at infinity. Since infinity is not defined, we can say that the curve never yields to a particular value at a very large value of ‘x’.

Complete step by step solution:
Now we start off with the solution to the given problem by writing that, we first need to find the value of the limit when ‘x’ approaches infinity. Thus we can write that,
  \[\displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\dfrac{\left( 3x-3 \right)}{\left( 4x+5 \right)}\]
Now, we divide both the numerator and the denominator by ‘x’ and from this we find,
\[\displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }\dfrac{\left( 3-\dfrac{3}{x} \right)}{\left( 4+\dfrac{5}{x} \right)}\]
Now if we put the value infinity in the place of ‘x’, the second term in the numerator and the denominator yields a value ‘0’. Thus,
\[\displaystyle \lim_{x \to \infty }f\left( x \right)=\dfrac{3}{4}\]
Therefore we can say that there is an horizontal asymptote at \[y=\dfrac{3}{4}\].

Note: For such types of problems we need to keep in mind the various formulae and equations of limits. We must also be very careful while evaluating the value of the limit as in some cases it may yield an incorrect answer. The horizontal asymptote must also be handled very cautiously as it decides the line of continuity or discontinuity.