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SOLUTION 28: a.) Start with −1≤sinx≤+1,
so that
3x−1≤3x+sinx≤3x+1
and (Assume that x>0.)
3x−12x≤3x+sinx2x≤3x+12x.
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 .
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