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Unformatted text preview: ∞ lim x→−∞ x2 + 1
|x | x2
+ | x|
1 + x2
+ |x |
|x | + 1
|x | 1 x2
= −1+0 = −1 = −1 Why did we multiply by 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.
Thus, we should multiply and divide by x. However, we want to bring x inside the square
x root sign. To do this we represent it as We know that 12
x is always positive. 2 (1)
But x is negative as x → −∞. Thus, if we have 1x we are multiplying by −1 instead of
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| 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
· 1 = lim
x→∞ 3 cos(ln(x)) − 5x2
− 2x + sin(πx)
x2 = lim 2
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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