Here you will learn how to solve or evaluate limits at infinity with examples.
Let’s begin –
Limits at Infinity
Algorithm to evaluate limits at infinity :
1). Write down the given expression in the form of a rational function i.e. \(f(x)\over g(x)\), if it is not so.
2). If k is the highest power of x in numerator and denominator both, then divide each term in numerator and denominator by \(x^k\).
3). Use the results \(\displaystyle{\lim_{x \to \infty}}\) \(c\over x^n\) = 0 and \(\displaystyle{\lim_{x \to \infty}}\) c = c, where n > 0.
Also Read : How to Solve Indeterminate Forms of Limits
Following examples will illustrate the above algorithm.
Example : Evaluate \(\displaystyle{\lim_{x \to \infty}}\) \(ax^2 + bx + c\over dx^2 + ex + f\).
Solution : Here the expression assumes the form \(\infty\over \infty\).
We notice that the highest power of x in both the numerator and denominator is 2.
So we divide each term in both the numerator and denominator by \(x^2\).
\(\therefore\) \(\displaystyle{\lim_{x \to \infty}}\) \(ax^2 + bx + c\over dx^2 + ex + f\)
= \(\displaystyle{\lim_{x \to \infty}}\) \(a + {b\over x} + {c\over x^2}\over d + {e\over x} + {f\over x^2}\)
= \(a + 0 + 0\over d + 0 + 0\) = \(a\over d\)
Example : Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\).
Solution : Here the expression assumes the form \(\infty\over \infty\).
We notice that the highest power of x in both the numerator and denominator is 2.
So we divide each term in both the numerator and denominator by \(x^2\).
\(\therefore\) \(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x – 5}\)
Limit = \(\displaystyle{\lim_{x \to 0}}\) \(1 + x + x^2\over {3 + 2x – 5x^2}\) = \(1\over 3\)
