Exercise 2.6 P.141
14. Evaluate the limit and justify each step by indicating the appropriate properties of limits.
xlim→∞
√ x3− 5x + 2 1 + 4x2+ 3x3 (sol)
∵√
x is continuous function
∴ limx
→∞
√
x3−5x+2 1+4x2+3x3 =√
xlim→∞
x3−5x+2 1+4x2+3x3 =
√
1 3
( lim
x→∞
x3−5x+2
1+4x2+3x3 = lim
x→∞
1−x25+ 2
x3 1
x3+4x+3 =13) 26. Find the limit or show that it does not exist.
x→−∞lim (x +√
x2+ 2x) (sol)
x→−∞lim (x +√
x2+ 2x) = lim
x→−∞(√x2+2x−x2
x2+2x−x) = lim
x→−∞( 2
−√
1+2x−1) =−1 27. Find the limit or show that it does not exist.
xlim→∞(√
x2+ ax−√
x2+ bx) (sol)
xlim→∞(√
x2+ ax−√
x2+ bx) = lim
x→∞((x√2+ax)−(x2+bx) x2+ax+√
x2+bx) = lim
x→∞(√ a−b
1+ax+√
1+bx) = a−b2 34. Find the limit or show that it does not exist.
xlim→∞
e3x− e−3x e3x+ e−3x (sol)
xlim→∞
e3x−e−3x e3x+e−3x = lim
x→∞
1−e−6x 1+e−6x = 1
37. Find the limit or show that it does not exist.
xlim→∞e−2xcos x (sol)
∵ lim
x→∞|e−2xcos x| ≤ lim
x→∞|e−2x| = 0
∴ lim
x→∞e−2xcos x = 0
38. Find the limit or show that it does not exist.
lim
x→0+tan−1(ln x) (sol)
Let t = ln x⇒ lim
x→0+tan−1(ln x) = lim
t→−∞tan−1t =−π2 44. Find the horizontal and vertical asymptotes of each curve.
y = 1 + x4 x2− x4 (sol)
1+x4
x2−x4 =−1 +x1+x2−x24 =−1 + x2(11+x−x)(1+x)2
⇒
{ horizontal asymptotes : y =−1
vertical asymptotes : x = 0, x = 1, x =−1
1
P.142
51. A function f is a ratio of quadratic functions and has a vertical asymptote x = 4 and just one x-intercept, x = 1. It is known that f has a removable discontinuity at x =−1 and lim
x→−1f (x) = 2.
Evaluate
(a) f (0) (b) lim
x→∞f (x) (sol)
f (x) = P (x)Q(x) and P (x), Q(x) are quadratic functions.
∵ f has vertical asymptote x = 4 and a removable discontinuity at x = −1
∴ Q(x) = a(x − 4)(x + 1) ∧ P (x) = (bx + c)(x + 1)
∵
{ f (1) = 0
xlim→−1f (x) = 2 ∴ { b+c
−3a= 0
−b+c−5a = 2 ⇒
{ b =−c
b a = 5 (a) f (0) = −4ac = 54
(b) lim
x→∞f (x) = lim
x→∞
(bx+c)(x+1)
a(x−4)(x+1) =ab = 5 59. Let P and Q be polynomials. Find
lim
x→∞
P (x) Q(x)
if the degree of P is (a) less than the degree of Q and (b) greater than the degree of Q.
(sol)
Let P (x) = anxn+ ... + a0 , Q(x) = bmxm+ ... + b0
(a) if m > n
P (x)
Q(x)= banxn+...+a0
mxm+...+b0 =b an+...+a0x−n
mxm−n+...+b0x−n ⇒ lim
x→∞
P (x) Q(x) = 0 (b) if m < n
P (x)
Q(x)= banxn+...+a0
mxm+...+b0 =anxbn−m+...+a0x−m
m+...+b0x−m ⇒ lim
x→∞
P (x) Q(x)=∞
60. Make a rough sketch of the curve y = xn (n an integer) for the following five case:
(i) n = 0 (ii) n > 0, n odd (iii) n > 0, n even (iv) n < 0 , n odd (v) n < 0, n even
Then use these sketches to find the following limits.
(a) lim
x→0+xn (b) lim
x→0−xn (c) lim
x→∞xn (d) lim
x→−∞xn (sol)
(i)(a) lim
x→0+xn = 1 (b) lim
x→0−xn= 1 (c) lim
x→∞xn= 1 (d) lim
x→−∞xn = 1
2
(ii)(a) lim
x→0+xn= 0 (b) lim
x→0−xn= 0 (c) lim
x→∞xn=∞ (d) lim
x→−∞xn=−∞
(iii)(a) lim
x→0+xn= 0 (b) lim
x→0−xn= 0 (c) lim
x→∞xn=∞ (d) lim
x→−∞xn=∞
(iv)(a) lim
x→0+xn =∞ (b) lim
x→0−xn=−∞ (c) lim
x→∞xn= 0 (d) lim
x→−∞xn= 0
(v)(a) lim
x→0+xn=∞ (b) lim
x→0−xn=∞ (c) lim
x→∞xn = 0 (d) lim
x→−∞xn= 0
3
