170
Chapter 2 | Limits
left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of x −3 x ( x −2) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator: lim x →2 − x −3 x 2 −2 x = lim x →2 − x −3 x ( x −2) . Step 2. Since x −2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/( x −2) from the rest of the function: = lim x →2 − x −3 x · 1 x −2 . Step3. lim x →2 − x −3 x = − 1 2 and lim x →2 − 1 x −2 =−∞. Therefore, the product of ( x −3)/ x and 1/( x −2) has a limit of +∞: lim x →2 − x −3 x 2 −2 x =+∞.
x +2 ( x −1) 2 .
2.18
Evaluate lim x →1
The Squeeze Theorem The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem , proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a . Figure 2.27 illustrates this idea.
Figure 2.27 The Squeeze Theorem applies when f ( x ) ≤ g ( x ) ≤ h ( x ) and lim x → a f ( x ) = lim x → a h ( x ).
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