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Unformatted text preview: ∞ lim x→−∞ x2 + 1
·
x+e 1
x2
1
x  x2
1
+ x2
x2
= lim
e
x
x→−∞
+  x
x 
1 1/2
1 + x2
e
x
+ x 
x 
1/2 1+
x
x  + 1
x2
e
x  1 x2
=
x
e
lim
+ lim
x→−∞ x
x→−∞ x
(1+0)1/2
1
= −1+0 = −1 = −1 Why did we multiply by 1
x
1
x lim 1 + lim x→−∞ x→−∞ ? Normally, we identify the largest power of x in the denominator and divide the
numerator and denominator by such. In this problem the largest power of x is x itself.
1
Thus, we should multiply and divide by x. However, we want to bring x inside the square
12
.
x root sign. To do this we represent it as We know that 12
x is always positive. 2 (1)
1
But x is negative as x → −∞. Thus, if we have 1x we are multiplying by −1 instead of
x
one, since the numerator is always positive and the denominator is always negative.
Therefore, to make sure that we are multiplying by 1, we multiply by
√
positive. We also know that x = x2 .
6. lim+ csc(x)= lim+ x→5π x→5π 1
x
1
x which is always 1
sin(x) We note that there is a vertical asymptote at x = 5π + . Then note that sin(x) < 0, when
5π < x < 6π . Thus, we conclude that limx→5π+ csc(x) = −∞. 2 1
3x2 − 2x + sin(πx) x2
7. lim
· 1 = lim
x→∞
x→∞ 3 cos(ln(x)) − 5x2
x2 3x2
− 2x + sin(πx)
x2
x2
x2
3 cos(ln(x))
5x2
− x2
x2 = lim 2
+ sin(πx)
x
x2
3 cos(ln(x))
−5
x2 3− x→∞ Recall that we can only use limit laws to evaluate the individual limits, if we are assured
that each of the limits, in fact, exists. Therefore, we must consider wh...
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 Summer '08
 FBHinkelmann
 Calculus, Limits

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