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Then for x > 1 we obtain0<xαxn< x.xaaUsing propertiesof the limit and the result of the preceding example, we find thatαlimx→+∞ xa x = 0.3.2 The Limit of a Function139Example 30 Let us show thata −1/x=0R+ x→0 x αlimfor a > 1 and any α ∈ R, that is, a −1/x = o(x α ) as x → 0, x ∈ R+ .Proof Setting x = −1/t in this case and using the theorem on the limit of a composite function and the result of the preceding example, we finda −1/xtα= lim t = 0.αt→+∞ aR+ x→0 xlimExample 31 Let us show thatloga x=0x→+∞ x αlimfor α > 0, that is, for any positive exponent α we have loga x = o(x α ) as x → +∞.Proof If a > 1, we set x = a t/α . Then by the properties of power functions andthe logarithm, the theorem on the limit of a composite function, and the result ofExample 29, we findlimx→+∞loga x(t/α) 1tlim t = 0.= lim=αtt→+∞t→+∞xaαaIf 0 < a < 1, then 1/a > 1, and after the substitution x = a −t/α , we obtainloga x(−t/α)t1lim= lim=−= 0.α−tx→+∞ xt→+∞ aα t→+∞ (1/a)tlimExample 32 Let us show further thatx α loga x = o(1)as x → 0, x ∈ R+for any α > 0.Proof We need to show that limR+ x→0 x α loga x = 0 for α > 0.
Setting x = 1/tand applying the theorem on the limit of a composite function and the result of thepreceding example, we findlimR+ x→0x α loga x = limt→+∞loga (1/t)loga t= − lim= 0.t→+∞ t αtαDefinition 23 Let us agree that the notation f =B O(g) or f = O(g) over the baseB (read “f is big-oh of g over B”) means that the relation f (x) = β(x)g(x) holdsultimately over B where β(x) is ultimately bounded over B.1403LimitsIn particular f =B O(1) means that the function f is ultimately bounded over B.Example 33 ( x1 + sin x)x = O(x) as x → ∞.Definition 24 The functions f and g are of the same order over B, and we writef " g over B, if f =B O(g) and g =B O(f ) simultaneously.Example 34 The functions (2 + sin x)x and x are of the same order as x → ∞, but(1 + sin x)x and x are not of the same order as x → ∞.The condition that f and g be of the same order over the base B is obviouslyequivalent to the condition that there exist c1 > 0 and c2 > 0 and an element B ∈ Bsuch that the relations c1 g(x) ≤ f (x) ≤ c2 g(x)hold on B, or, what is the same, 1 1 f (x) ≤ g(x) ≤ f (x).c2c1Definition 25 If the relation f (x) = γ (x)g(x) holds ultimately over B wherelimB γ (x) = 1, we say that the function f behaves asymptotically like g over B,or, more briefly, that f is equivalent to g over B.In this case we shall write f ∼B g or f ∼ g over B.The use of the word equivalent is justified by the relationsf ∼f ,Bf ∼g ⇒ g∼f ,BB f ∼g ∧ g∼h ⇒ f ∼h .BBBIndeed, the relation f ∼B f is obvious, since in this case γ (x) ≡ 1.
Next, if11= 1 and g(x) = γ (x)f (x). Here all we that need tolimB γ (x) = 1, then limB γ (x)explain is why it is permissible to assume that γ (x) = 0. If the relation f (x) =γ (x)g(x) holds on B1 ∈ B, and 12 < |γ (x)| < 32 on B2 ∈ B, then we can take B ∈ Bwith B ⊂ B1 ∩ B2 , on which both relations hold. Outside of B, if convenient, wemay assume that γ (x) ≡ 1. Thus we do indeed have (f ∼ g) ⇒ (g ∼ f ).Finally, if f (x) = γ1 (x)g(x) on B1 ∈ B and g(x) = γ2 (x)h(x) on B2 ∈ B, thenon an element B ∈ B such that B ⊂ B1 ∩ B2 , both of these relations hold simultaneously, and so f (x) = γ1 (x)γ2 (x)h(x) on B.
But limB γ1 (x)γ2 (x) = limB γ1 (x) ·limB γ2 (x) = 1, and hence we have verified that f ∼B h.3.2 The Limit of a Function141It is useful to note that since the relation limB γ (x) = 1 is equivalent to γ (x) =1 + α(x), where limB α(x) = 0, the relation f ∼B g is equivalent to f (x) = g(x) +α(x)g(x) = g(x) + o(g(x)) over B.We see that the relative error |α(x)| = | f (x)−g(x)| in approximating f (x) by ag(x)function g(x) that is equivalent to f (x) over B is infinitesimal over B.Let us now consider some examples.Example 35 x 2 + x = (1 + x1 )x 2 ∼ x 2 as x → ∞.The absolute value of the difference of these functions 2 x + x − x 2 = |x|1tends to infinity.
However, the relative error |x|= |x|that results from replacingx222x + x by the equivalent function x tends to zero as x → ∞.Example 36 At the beginning of this discussion we spoke of the famous asymptoticlaw of distribution of the prime numbers. We can now give a precise statement ofthis law:xx+oas x → +∞.π(x) =ln xln xExample 37 Since limx→0 sinx x = 1, we have sin x ∼ x as x → 0, which can also bewritten as sin x = x + o(x) as x → 0.Example 38 Let us show that ln(1 + x) ∼ x as x → 0.Proofln(1 + x)= lim ln(1 + x)1/x = ln lim (1 + x)1/x = ln e = 1.x→0x→0x→0xlimHere we have used the relation loga (bα ) = α loga b in the first equality and therelation limt→b loga t = loga b = loga (limt→b t) in the second.Thus, ln(1 + x) = x + o(x) as x → 0.Example 39 Let us show that ex = 1 + x + o(x) as x → 0.Proofex − 1t= lim= 1.x→0t→0 ln(1 + t)xlimHere we have made the substitution x = ln(1 + t), ex − 1 = t and used the relationsex → e0 = 1 as x → 0 and ex = 1 for x = 0.
Thus, using the theorem on the limit of1423Limitsa composite function and the result of the preceding example, we have proved theassertion.Thus, ex − 1 ∼ x as x → 0.Example 40 Let us show that (1 + x)α = 1 + αx + o(x) as x → 0.Proof(1 + x)α − 1eα ln(1+x) − 1 α ln(1 + x)= lim.=x→0x→0 α ln(1 + x)xxlimet − 1ln(1 + x)· lim= α.t→0x→0tx= α limIn this computation, assuming α = 0, we made the substitution α ln(1 + x) = t andused the results of the two preceding examples.If α = 0, the assertion is obvious.Thus, (1 + x)α − 1 ∼ αx as x → 0.The following simple fact is sometimes useful in computing limits.Proposition 3 If f ∼B f˜, then limB f (x)g(x) = limB f˜(x)g(x), provided one ofthese limits exists.Proof Indeed, given that f (x) = γ (x)f˜(x) and limB γ (x) = 1, we havelim f (x)g(x) = lim γ (x)f˜(x)g(x) = lim γ (x) · lim f˜(x)g(x) = lim f˜(x)g(x).
BBBBBExample 411ln cos xln cos2 x 1ln(1 − sin2 x)limlim===x→0 sin(x 2 )2 x→0 x 22 x→0x2lim=1− sin2 xx211limlim=−=− .2 x→0 x 22 x→0 x 22Here we have used the relations ln(1 + α) ∼ α as α → 0, sin x ∼ x as x → 0,1122sin β ∼ β as β → 0, and sin x ∼ x as x → 0.We have proved that one may replace functions by other functions equivalent tothem in a given base when computing limits of monomials.
This rule should not beextended to sums and differences of functions.Example 42√x 2 + x ∼ x as x → +∞, but#x 2 + x − x = lim (x − x) = 0.limx→+∞x→+∞3.2 The Limit of a Function143In fact,limx→+∞#x11= lim = .x 2 + x − x = lim √x→+∞ x 2 + x + xx→+∞11+ x +1 2We note one more widely used rule for handling the symbols o(·) and O(·) inanalysis.Proposition 4 For a given basea)b)c)d)e)o(f ) + o(f ) = o(f );o(f ) is also O(f );o(f ) + O(f ) = O(f );O(f ) + O(f ) = O(f );(x))f (x)if g(x) = 0, then o(fg(x) = o( g(x) ) andO(f (x))g(x)(x)= O( fg(x)).Notice some peculiarities of operations with the symbols o(·) and O(·) thatfollow from the meaning of these symbols. For example 2o(f ) = o(f ) ando(f ) + O(f ) = O(f ) (even though in general o(f ) = 0); also, o(f ) = O(f ), butO(f ) = o(f ). Here the equality sign is used in the sense of “is”. The symbols o(·)and O(·) do not really denote a function, but rather indicate its asymptotic behavior,a behavior that many functions may have simultaneously, for example, f and 2f ,and the like.Proof a) After the clarification just given, this assertion ceases to appear strange.The first symbol o(f ) in it denotes a function of the form α1 (x)f (x), wherelimB α1 (x) = 0.
The second symbol o(f ), which one can (or should) equip withsome mark to distinguish it from the first, denotes a function of the form α2 (x)f (x),where limB α2 (x) = 0. Then α1 (x)f (x) + α2 (x)f (x) = (α1 (x) + α2 (x))f (x) =α3 (x)f (x), where limB α3 (x) = 0.Assertion b) follows from the fact that any function having a limit is ultimatelybounded.Assertion c) follows from b) and d).Assertion d) follows from the fact that the sum of ultimately bounded functionsis ultimately bounded.(x))α(x)f (x)f (x)f (x)As for e), we have o(fg(x) = g(x) = α(x) g(x) = o( g(x) ).The second part of assertion e) is verified similarly.Using these rules and the equivalences obtained in Example 40, we can now findthe limit in Example 42 by the following direct method:limx + x − x = lim x1+ −1 =x→+∞x→+∞x 11 1−1 == lim x 1 + · + ox→+∞2 xx1443Limits 11+x·o=x→+∞ 2x11+ o(1) = .= limx→+∞ 22= limWe shall soon prove the following important relations, which should be memorized at this point like the multiplication table:111x + x2 + · · · + xn + · · ·1!2!n!11(−1)k 2kx + ···cos x = 1 − x 2 + x 4 + · · · +2!4!(2k)!ex = 1 +sin x =1(−1)k 2k+11+ ···x − x3 + · · · +x1!3!(2k + 1)!for x ∈ R,for x ∈ R,for x ∈ R,11(−1)n−1 nln(1 + x) = x − x 2 + x 3 + · · · +x + · · · for |x| < 1,23nα(α − 1) 2α(1 + x)α = 1 + x +x + ···+1!2!α(α − 1) · · · (α − n + 1) nx + ···+for |x| < 1.n!On the one hand, these relations can already be used as computational formulas, andon the other hand they contain the following asymptotic formulas, which generalizethe formulas contained in Examples 37–40:111x + x 2 + · · · + x n + O x n+11!2!n!11(−1)k 2kcos x = 1 − x 2 + x 4 + · · · +x + O x 2k+22!4!(2k)!ex = 1 +sin x =as x → 0,as x → 0,11(−1)k 2k+1x − x3 + · · · +x+ O x 2k+3 as x → 0,1!3!(2k + 1)!11(−1)n−1 nln(1 + x) = x − x 2 + x 3 + · · · +x + O x n+123nα(α−1)α(1 + x)α = 1 + x +x2 + · · · +1!2!α(α − 1) · · · (α − n + 1) nx + O x n+1+n!as x → 0,as x → 0.These formulas are usually the most efficient method of finding the limits of theelementary functions.
When doing so, it is useful to keep in mind that O(x m+1 ) =x m+1 · O(1) = x m · xO(1) = x m o(1) = o(x m ) as x → 0.In conclusion, let us consider a few examples showing these formulas in action.3.2 The Limit of a Function145Example 43 2x − (x − 3!1 x 3 + O(x 5 ))1x − sin x1+O xlim= .= lim= limx→0x→0x→0 3!3!x3x3Example 44 Let us findlim x2x→∞'7x3 + x1.− cosx1 + x3As x → ∞ we have:'7x3 + x1 −11 + x −211+ 3== 1+ 2=1 + x31 + x −3xx 111111− 3 +O 6=1+ 2 +O 3 ,= 1+ 2xxxxx 1/7 x3 + x111 11= 1+ 2 +O 3=1+ · 2 +O 3 ,7 x1 + x3xxx 11 11cos = 1 − · 2 + O 4 ,x2! xxfrom which we obtain' 39 1117 x +xas x → ∞.·− cos =+O 3x 14 x 21 + x3xHence the required limit islim xx→∞2 991= .+O 321414xxExample 45$ % 11 x x1 xlim1+= lim exp x ln 1 +−1=x→∞ ex→∞xx1−x == lim exp x 2 ln 1 +x→∞x 111= lim exp x 2−x =− 2 +O 3x→∞x 2xx 11= e−1/2 .= lim exp − + Ox→∞2x1463Limits3.2.5 Problems and Exercises1.
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