2.6 - Review Key

# In this problem the largest power of x is x itself 1

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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) &lt; 0, when 5π &lt; x &lt; 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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