Properties of Limits
Solved Problems
Example 3.
Suppose that \(\lim\limits_{x \to 1} f\left( x \right) = 2\) and \(\lim\limits_{x \to 1} g\left( x \right) = 3.\) Calculate the limit \[\lim\limits_{x \to 1} {\frac{{g\left( x \right) - 3f\left( x \right)}}{{{f^2}\left( x \right) + g\left( x \right)}}}.\]
Solution.
\[\lim\limits_{x \to 1} \frac{{g\left( x \right) - 3f\left( x \right)}}{{{f^2}\left( x \right) + g\left( x \right)}} = \frac{{\lim\limits_{x \to 1} \left[ {g\left( x \right) - 3f\left( x \right)} \right]}}{{\lim\limits_{x \to 1} \left[ {{f^2}\left( x \right) + g\left( x \right)} \right]}} = \frac{{\lim\limits_{x \to 1} g\left( x \right) - \lim\limits_{x \to 1} \left[ {3f\left( x \right)} \right]}}{{\lim\limits_{x \to 1} {f^2}\left( x \right) + \lim\limits_{x \to 1} g\left( x \right)}} = \frac{{\lim\limits_{x \to 1} g\left( x \right) - 3\lim\limits_{x \to 1} f\left( x \right)}}{{{{\left[ {\lim\limits_{x \to 1} f\left( x \right)} \right]}^2} + \lim\limits_{x \to 1} g\left( x \right)}} = \frac{{3 - 3 \cdot 2}}{{{2^2} + 3}} = - \frac{3}{7}.\]
Example 4.
Calculate the limit \[\lim\limits_{x \to \infty } {\frac{{3x + \cos x}}{{2x - 7}}}.\]
Solution.
We know that \( - 1 \le \cos x \le 1\) for all \(x.\) Then
\[3x - 1 \le 3x + \cos x \le 3x + 1.\]
Dividing by \(2x - 7 \gt 0,\) we obtain
\[\frac{{3x - 1}}{{2x - 7}} \le \frac{{3x + \cos x}}{{2x - 7}} \le \frac{{3x + 1}}{{2x - 7}}.\]
Since we consider large and positive \(x,\) and, hence, \(2x - 7 \gt 0,\) we do not switch the inequality signs. Then
\[\lim\limits_{x \to \infty } \frac{{3x - 1}}{{2x - 7}} \le \lim\limits_{x \to \infty } \frac{{3x + \cos x}}{{2x - 7}} \le \lim\limits_{x \to \infty } \frac{{3x + 1}}{{2x - 7}}.\]
Calculate the left and right limits:
\[\lim\limits_{x \to \infty } \frac{{3x - 1}}{{2x - 7}} = \lim\limits_{x \to \infty } \frac{{3 - \frac{1}{x}}}{{2 - \frac{7}{x}}} = \frac{3}{2},\;\;\; \lim\limits_{x \to \infty } \frac{{3x + 1}}{{2x - 7}} = \lim\limits_{x \to \infty } \frac{{3 + \frac{1}{x}}}{{2 - \frac{7}{x}}} = \frac{3}{2}.\]
It follows from the Squeeze Theorem that
\[\lim\limits_{x \to \infty } \frac{{3x + \cos x}}{{2x - 7}} = \frac{3}{2}.\]
Example 5.
Find the limit \[\lim\limits_{x \to \infty } {\frac{{2\sin x - 5x}}{{3x + 1}}}.\]
Solution.
We know that \( - 1 \le \sin x \le 1\) for all \(x.\) Then
\[ - 2 \le 2\sin x \le 2.\]
Subtract \(5x\) from all terms of the inequality.
\[ - 2 - 5x \le 2\sin x - 5x \le 2 - 5x.\]
Dividing by \(3x + 1,\) we get
\[\frac{{ - 2 - 5x}}{{3x + 1}} \le \frac{{2\sin x - 5x}}{{3x + 1}} \le \frac{{2 - 5x}}{{3x + 1}}.\]
(We do not switch the inequality signs, since \(3x + 1\) is positive as \(x \to \infty.\))
Calculate the left and right limits:
\[\lim\limits_{x \to \infty } \frac{{ - 2 - 5x}}{{3x + 1}} = \lim\limits_{x \to \infty } \frac{{ - \frac{2}{x} - 5}}{{3 + \frac{1}{x}}} = - \frac{5}{3},\;\;\; \lim\limits_{x \to \infty } \frac{{2 - 5x}}{{3x + 1}} = \lim\limits_{x \to \infty } \frac{{\frac{2}{x} - 5}}{{3 + \frac{1}{x}}} = - \frac{5}{3}.\]
We see that the left and right limits are equal to each other. Then by the Squeeze Theorem,
\[\lim\limits_{x \to \infty } \frac{{2\sin x - 5x}}{{3x + 1}} = - \frac{5}{3}.\]
