SOLUTION 28: a.) Start with $$ \displaystyle{ -1 \le \sin x \le +1 } ,$$
so that
$$ \displaystyle{ 3x-1 \le 3x+\sin x \le 3x+1 } $$
and (Assume that $ x>0 $.)
$$ \displaystyle{ { 3x-1 \over 2x } \le { 3x+\sin x \over 2x } \le { 3x+1 \over 2x } } .$$
Then
$$ \displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } }
= \displaystyle{ `` \infty "\over \infty } = \displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } \cdot { 1/x \over 1/x } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \ { 3x/x -1/x \over 2x/x } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \ { 3x/x -1/x \over 2x/x } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \ { 3 - \frac{``1"}{\infty} \over 2 } } $$
$$ = \displaystyle{ 3 - 0 \over 2 } $$
$$ = \displaystyle{ 3 \over 2 } .$$
In a similar fashion it can easily be shown that
$$ \displaystyle{ \lim_{x \to \infty} \ { 3x+1 \over 2x } }
= \displaystyle{ 3 \over 2 } .$$
Since
$$ \displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } }
= \displaystyle{ 3 \over 2 } = \displaystyle{ \lim_{x \to \infty} \ { 3x+1 \over 2x } } , $$
it follows from the Squeeze Principle that
$$ \displaystyle{ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } }
= \displaystyle{ 3 \over 2 } .$$
SOLUTION 28: b.) $$ \displaystyle{ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } }
= \displaystyle{ `` \ 3 \cdot \infty + (some \ number \ between \ -1 \ and \ +1) \ " \over 2 \cdot \infty }
= \displaystyle{ `` \ \infty + (some \ number \ between \ -1 \ and \ +1) \ " \over \infty }
= \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule.)
$$ = \displaystyle{ \lim_{x \to \infty} \ { 3+ \cos x \over 2 } } , $$
which DOES NOT EXIST since
$$ \displaystyle{ -1 \le \cos x \le +1 } $$
SOLUTION 28: c.) The answers to parts a.) and b.) tell us that l'Hopital's Rule may give us a wrong answer if the answer is `` does not exist." We can only be sure that l'Hopital's Rule gives us the correct answer if the answer is finite, $ + \infty $, or $ - \infty $ .
Click HERE to return to the list of problems.