SOLUTIONS TO LIMITS USING THE SQUEEZE PRINCIPLE

SOLUTION 1 : First note that

$-1 \le \sin x \le +1$

because of the well-known properties of the sine function. Since we are computing the limit as x goes to infinity, it is reasonable to assume that x > 0 . Thus,

$${ { -1 \over x } \le { \sin x \over x } \le { 1 \over x } }$$ .

Since

$${ \lim_{ x \to \infty } \ { -1 \over x } = 0 = \lim_{ x \to \infty } \ { 1 \over x } }$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ x \to \infty } \ { \sin x \over x } = 0 }$$ .

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SOLUTION 2 : First note that

$-1 \le \cos x \le +1$

because of the well-known properties of the cosine function. Now multiply by -1, reversing the inequalities and getting

$+1 \ge - \cos x \ge -1$

or

$-1 \le - \cos x \le +1$ .

Next, add 2 to each component to get

$1 \le 2 - \cos x \le 3$ .

Since we are computing the limit as x goes to infinity, it is reasonable to assume that x + 3 > 0. Thus,

$${ { 1 \over x+3 } \le { 2 - \cos x \over x+3 } \le { 3 \over x+3 } }$$ .

Since

$${ \lim_{ x \to \infty } \ { 1 \over x+3 } = 0 = \lim_{ x \to \infty } \ { 3 \over x+3 } }$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ x \to \infty } \ { 2 - \cos x \over x+3 } = 0 }$$ .

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SOLUTION 3 : First note that

$-1 \le \cos(2x)\le +1$

because of the well-known properties of the cosine function, and therefore

$0 \le \cos^2 (2x)\le +1$ .

Since we are computing the limit as x goes to infinity, it is reasonable to assume that 3 - 2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting

$${ { 0 \over 3-2x } \ge { \cos^2 (2x) \over 3-2x } \ge { 1 \over 3-2x } }$$ ,

or

$${ { 1 \over 3-2x } \le { \cos^2 (2x) \over 3-2x } \le 0 }$$ .

Since

$${ \lim_{ x \to \infty } \ { 1 \over 3-2x } = 0 = \lim_{ x \to \infty } \ 0 }$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ x \to \infty } \ { \cos^2 (2x) \over 3-2x } = 0 }$$ .

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SOLUTION 4 : Note that $${ \lim_{ x \to 0^{-} } \ \cos \Big( { 2 \over x }\Big) }$$ DOES NOT EXIST since values of $${ \cos \Big( { 2 \over x }\Big) }$$ oscillate between -1 and +1 as x approaches 0 from the left. However, this does NOT necessarily mean that $${ \lim_{ x \to 0^{-} } \ { x^3 \cos \Big( { 2 \over x }\Big) } }$$ does not exist ! ? #. Indeed, x³ < 0 and

$${ -1 \le \cos \Big( { 2 \over x }\Big) \le +1 }$$

for x < 0. Multiply each component by x³, reversing the inequalities and getting

$${ -x^3 \ge x^3 \cos \Big( { 2 \over x }\Big) \ge x^3 }$$

or

$${ x^3 \le x^3 \cos \Big( { 2 \over x }\Big) \le -x^3 }$$ .

Since

$${ \lim_{ x \to 0^{-} } \ x^3 = 0 = \lim_{ x \to 0^{-} } \ \{-x^3 \} }$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ x \to 0^{-} } \ x^3 \cos \Big( { 2 \over x }\Big) = 0 }$$ .

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SOLUTION 5 : First note that

$-1 \le \sin x \le +1$ ,

so that

$0 \le \sin^2 x \le 1$

and

$2 \le 2 + \sin^2 x \le 3$ .

Since we are computing the limit as x goes to infinity, it is reasonable to assume that x+100 > 0. Thus, dividing by x+100 and multiplying by x², we get

$${ { 2 \over x + 100 } \le { 2 + \sin^2 x \over x + 100 } \le { 3 \over x + 100 } }$$

and

$${ { 2 x^2\over x + 100 } \le { x^2 (2 + \sin^2 x ) \over x + 100 } \le { 3x^2 \over x + 100 } }$$ .

Then

$${ \lim_{ x \to \infty } \ { 2x^2 \over x+100 } }$$ = $${ \lim_{ x \to \infty } \ { 2x^2 \over x+100 } \ { \displaystyle{ 1 \over x } \over \displaystyle{ 1 \over x } } }$$

= $${ \lim_{ x \to \infty } \ { 2x \over 1 + \displaystyle{100 \over x } } }$$

= $${ \infty \over 1 + 0 }$$

= $\infty$ .

Similarly,

$${ \lim_{ x \to \infty } \ { 3x^2 \over x+100 } }$$ = $\infty$ .

Thus, it follows from the Squeeze Principle that

$${ \lim_{ x \to \infty } \ { x^2 (2 + \sin^2 x) \over x+100 } }$$ = $\infty$ (does not exist).

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SOLUTION 6 : First note that

$-1 \le \sin(3x) \le +1$ ,

so that

$-1 \le - \sin(3x) \le +1$ ,

$5x^2 -1 \le 5x^2 - \sin(3x) \le 5x^2 + 1$ ,

and

$${ { 5x^2-1 \over x^2 + 10 } \le { 5x^2 - \sin(3x) \over x^2 + 10 } \le { 5x^2 +1 \over x^2 + 10 } }$$ .

Then

$${ \lim_{ x \to -\infty } \ { 5x^2 - 1 \over x^2+10 } }$$

= $${ \lim_{ x \to -\infty } \ { 5x^2 - 1 \over x^2+10 } \ { \displaystyle{ 1 \over x^2 } \over \displaystyle{ 1 \over x^2 } } }$$

= $${ \lim_{ x \to -\infty } \ { 5 - \displaystyle{ 1 \over x^2 } \over 1 + \displaystyle{ 10 \over x^2 } } }$$

= $${ 5 - 0 \over 1 + 0 }$$

= 5 .

Similarly,

$${ \lim_{ x \to -\infty } \ { 5x^2 + 1 \over x^2+10 } }$$ = 5 .

Thus, it follows from the Squeeze Principle that

$${ \lim_{ x \to -\infty } \ { 5x^2 - \sin(3x) \over x^2+10 } }$$ = 5 .

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SOLUTION 7 : First note that

$-1 \le \sin x \le +1$

and

$-1 \le \cos x \le +1$ ,

so that

$-1 \le \cos^3 x \le +1$

and

$-2 \le \sin x + \cos^3 x \le +2$ .

Since we are computing the limit as x goes to negative infinity, it is reasonable to assume that x-3 < 0. Thus, dividing by x-3, we get

$${ { -2 \over x-3 } \ge { \sin x + \cos^3 x \over x-3 } \ge { 2 \over x-3 } }$$

or

$${ { 2 \over x-3 } \le { \sin x + \cos^3 x \over x-3 } \le { -2 \over x-3 } }$$ .

Now divide by x² + 1 and multiply by x² , getting

$${ { 2 x^2\over (x^2 + 1) (x-3) } \le { x^2 (\sin x + \cos^3 x ) \over (x^2 + 1) (x-3) } \le { -2x^2 \over (x^2 + 1) (x-3) } }$$ .

Then

$${ \lim_{ x \to -\infty } \ { 2 x^2 \over (x^2+1) (x-3) } }$$

= $${ \lim_{ x \to -\infty } \ { 2 x^2 \over x^3 - 3x^2 + x - 3 } }$$

= $${ \lim_{ x \to -\infty } \ { 2 x^2 \over x^3 - 3x^2 + x - 3 } \ { \displaystyle{ 1 \over x^3 } \over \displaystyle{ 1 \over x^3 } } }$$

= $${ \lim_{ x \to -\infty } \ { \displaystyle{ 2 \over x } \over 1 -... ... 3 \over x } + \displaystyle{ 1 \over x^2 } - \displaystyle{ 3 \over x^3 } } }$$

= $${ 0 \over 1 - 0 + 0 - 0 }$$

= 0 .

Similarly,

$${ \lim_{ x \to -\infty } \ { -2 x^2 \over (x^2+1) (x-3) } }$$ = 0 .

It follows from the Squeeze Principle that

$${ \lim_{ x \to -\infty } \ { x^2 ( \sin x + \cos^3 x ) \over (x^2+1) (x-3) } }$$ = 0 .

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SOLUTION 8 : Since

$${ \lim_{ \theta \to {-1^{+} } } { \theta^2 + \theta - 2 \over \theta + 3 } }$$ = $${ { (-1)^2 + (-1) - 2 \over (-1) + 3 } } = -1$$

and

$${ \lim_{ \theta \to {-1^{+} } } { \theta^2 + 2 \theta - 1 \over \theta + 3 } }$$ = $${ { (-1)^2 + 2(-1) - 1 \over (-1) + 3 } } = -1$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ \theta \to {-1^{+} } } { f( \theta ) \over \theta^2 } } = -1$$ ,

that is,

$-1 = \displaystyle{ \lim_{ \theta \to {-1^{+} } } { f( \theta ) \over \theta^2 } }$

$= \displaystyle{ \displaystyle{ \lim_{ \theta \to {-1^{+} } } f( \theta ) } \over \displaystyle{ \lim_{ \theta \to {-1^{+} } } \theta^2 } }$

$= \displaystyle{ \displaystyle{ \lim_{ \theta \to {-1^{+} } } f( \theta ) } \over (-1)^2 }$

$= \displaystyle{ \lim_{ \theta \to {-1^{+} } } f( \theta ) }$ .

Thus,

$${ \lim_{ \theta \to {-1^{+} } } f( \theta ) } = -1$$ .

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SOLUTION 9 : a.) First note that (See diagram below.)

area of triangle OAD < area of sector OAC < area of triangle OBC .

The area of triangle OAD is

$${ 1 \over 2 }$$ (base) (height) $= \displaystyle{ 1 \over 2 } (\cos \theta) (\sin \theta)$ .

The area of sector OAC is

$${ \theta \over 2 \pi }$$ (area of circle) $= \displaystyle{ \theta \over 2 \pi } (\pi (1)^2 ) = \displaystyle{ \theta \over 2 }$ .

The area of triangle OBC is

$${ 1 \over 2 }$$ (base) (height) $= \displaystyle{ 1 \over 2 } (1) (\tan \theta) = \displaystyle{ 1 \over 2 } \displaystyle{ \sin \theta \over \cos \theta }$ .

It follows that

$${ 1 \over 2 } (\cos \theta) (\sin \theta) < \displaystyle{ \theta... ... < \displaystyle{ 1 \over 2 } \displaystyle{ \sin \theta \over \cos \theta }$$

or

$(\cos \theta) (\sin \theta) < \theta < \displaystyle{ \sin \theta \over \cos \theta }$ .

b.) If $0 < \theta < \displaystyle{ \pi \over 2 }$, then $\sin \theta > 0$ and $\cos \theta > 0$ , so that dividing by $\sin \theta$ results in

$\cos \theta < \displaystyle{ \theta \over \sin \theta } < \displaystyle{ 1 \over \cos \theta }$ .

Taking reciprocals of these positive quantities gives

$${ 1 \over \cos \theta } > \displaystyle{ \sin \theta \over \theta } > \cos \theta$$

or

$\cos \theta < \displaystyle{ \sin \theta \over \theta } < \displaystyle{ 1 \over \cos \theta }$ .

Since

$${ \lim_{ \theta \to 0^{+} } \ { \cos \theta } } = \cos 0 = 1 = \displaystyle{ \lim_{ \theta \to 0^{+} } \ \Big( { 1 \over \cos \theta } \Big) }$$ ,

it follows from the Squeeze Principle that

$${ \lim_{ \theta \to 0^{+} } \ { \sin \theta \over \theta } } = 1$$ .

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SOLUTION 10 : Recall that function f is continuous at x=0 if

i.) f(0) is defined ,

ii.) $${ \lim_{ x \to 0 } f(x) }$$ exists ,

and

iii.) $${ \lim_{ x \to 0 } f(x) = f(0) }$$ .

First note that it is given that

i.) f(0) = 0 .

Use the Squeeze Principle to compute $${ \lim_{ x \to 0 } f(x) }$$ . For $x \ne 0$ we know that

$-1 \le \sin \Big( \displaystyle{ 1 \over x } \Big) \le +1$ ,

so that

$-x^2 \le x^2 \sin \Big( \displaystyle{ 1 \over x } \Big) \le x^2$ .

Since

$${ \lim_{ x \to 0 } (-x^2) = 0 = \lim_{ x \to 0 } x^2 }$$

it follows from the Squeeze Principle that

ii.) $${ \lim_{ x \to 0 } f(x) } = \displaystyle{ \lim_{ x \to 0 } \ x^2 \sin \Big( \displaystyle{ 1 \over x } \Big) } = 0$$ .

Finally,

iii.) $${ \lim_{ x \to 0 } f(x) = 0 = f(0) }$$ ,

confirming that function f is continuous at x=0 .

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