Limits Using Taylor Expansions: Original Solved Exercises
Exercise editors and source: Prof. Antonino De Martino and Dr. Luana Manfredini, Eserciziario 2.1. The original exercises and source theory are presented here in English for Logic & Motion.
Taylor and Maclaurin Expansions
Taylor expansions are useful because they replace transcendental functions by polynomial expressions, with a remainder that records the approximation order. A sufficient hypothesis is that f belongs to Cⁿ on a neighborhood of x₀. As x → x₀, Taylor’s formula with Peano remainder is:
\[f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 +\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n).\]At x₀ = 0 this is called a Maclaurin expansion.
Algebra of Little-o Terms
The following relations concern x → 0; logarithms are natural, and trigonometric arguments are in radians.
\[o(x^m)+o(x^m)=o(x^m).\] \[o(x^m)o(x^n)=o(x^{m+n}).\] \[o(x^n)+o(x^m)=o(x^{\min\{m,n\}}).\] \[x^n o(x^m)=o(x^{n+m}).\] \[o(x^n+o(x^n))=o(x^n).\]The last identity is the composition rule used when the argument of a remainder itself has an asymptotic expansion.
Maclaurin Formulas
\[(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2 +\cdots+\binom{\alpha}{n}x^n+o(x^n).\] \[e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+o(x^n).\]For a > 0:
\[a^x=1+x\log a+\frac{x^2}{2!}\log^2 a +\cdots+\frac{x^n}{n!}\log^n a+o(x^n).\] \[\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3} +\cdots+(-1)^{n+1}\frac{x^n}{n}+o(x^n).\] \[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!} +\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+2}).\] \[\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!} +\cdots+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1}).\] \[\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15} +\frac{17x^7}{315}+\frac{62x^9}{2835}+o(x^{10}).\] \[\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5} +\cdots+(-1)^n\frac{x^{2n+1}}{2n+1}+o(x^{2n+2}).\] \[\arcsin x=x+\frac{x^3}{6}+\frac{3x^5}{40} +\cdots+\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}+o(x^{2n+2}).\] \[\arccos x=\frac\pi2-x-\frac{x^3}{6}-\frac{3x^5}{40} -\cdots-\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}+o(x^{2n+2}).\] \[\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!} +\cdots+\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+2}).\] \[\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!} +\cdots+\frac{x^{2n}}{(2n)!}+o(x^{2n+1}).\]L’Hôpital’s Theorem
Let f and g be defined on a punctured neighborhood of c. Suppose both tend to zero, or both are infinite, and both are differentiable there with g′ nonzero. If the following limit exists, finite or infinite:
\[\lim_{x\to c}\frac{f'(x)}{g'(x)}=L,\]then:
\[\lim_{x\to c}\frac{f(x)}{g(x)}=L.\]The theorem also applies to one-sided limits and limits at infinity under the corresponding hypotheses.
Worked Exercises
Fifteen worked exercises apply Taylor expansions to limits with cancellations at different orders.
Exercise 1
Evaluate the following limit:
\[\lim_{x \to 0}\frac{\sin x -\frac{1}{2}x+x^{3}}{\tan x -1+ \cos x}.\]Solution.
Expand numerator and denominator to third order: \(\sin x-\frac x2+x^3=\frac x2+\frac56x^3+o(x^3).\) \(\tan x-1+\cos x=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3).\) Factor x in both expressions: \(\frac{\frac12+\frac56x^2+o(x^2)}{1-\frac x2+\frac{x^2}{3}+o(x^2)}\longrightarrow\frac12.\)
Final result
\[\frac12\]Exercise 2
Evaluate the following limit:
\[\lim_{x \to 0}\frac{\sinh x -\sin x}{x^{3}}.\]Solution.
The linear terms cancel: \(\sinh x-\sin x =\left(x+\frac{x^3}{6}+o(x^3)\right) -\left(x-\frac{x^3}{6}+o(x^3)\right).\) \(\frac{\sinh x-\sin x}{x^3}=\frac{x^3/3+o(x^3)}{x^3}\longrightarrow\frac13.\)
Final result
\[\frac13\]Exercise 3
Evaluate the following limit:
\[\lim_{x \to 0}\frac{(2+\cos(3x)-3 \cosh x)^{4}}{\log(1+x^{2})}.\]Solution.
Use the quadratic terms before taking the fourth power: \(2+\cos(3x)-3\cosh x =2+1-\frac92x^2-3-\frac32x^2+o(x^2) =-6x^2+o(x^2).\) \(\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4).\) \(\frac{(-6x^2+o(x^2))^4}{x^2-x^4/2+o(x^4)} =\frac{6^4x^8+o(x^8)}{x^2(1-x^2/2+o(x^2))}\longrightarrow0.\)
Final result
\[0\]Exercise 4
Evaluate the following limit:
\[\lim_{x \to 0}(\log(1+x)+\cos ^{2}x)^{\frac{1}{x}}\]Solution.
The base is positive near zero. Write the expression as an exponential: \((\log(1+x)+\cos^2x)^{1/x} =\exp\!\left(\frac{\log(\log(1+x)+\cos^2x)}{x}\right).\) \(\log(1+x)+\cos^2x=1+x+o(x).\) The composition rule for little-o terms gives: \(\log(1+x+o(x))=x+o(x)+o(x+o(x))=x+o(x).\) \(\frac{\log(\log(1+x)+\cos^2x)}x\longrightarrow1.\)
Final result
\[e\]Exercise 5
Evaluate the following limit:
\[\lim_{x \to 1^{+}}\frac{\sin (\sqrt{x-1})-\sqrt{\log x}}{x-1}.\]Solution.
Set h = x − 1 > 0. Expand the two radicals to the same order: \(\sin\sqrt h=\sqrt h-\frac{h^{3/2}}6+o(h^{3/2}).\) \(\log(1+h)=h-\frac{h^2}{2}+o(h^2).\) \(\sqrt{\log(1+h)}=\sqrt h\left(1-\frac h2+o(h)\right)^{1/2}.\) \(\sqrt{\log(1+h)}=\sqrt h\left(1-\frac h4+o(h)\right) =\sqrt h-\frac{h^{3/2}}4+o(h^{3/2}).\) \(\frac{\sin\sqrt h-\sqrt{\log(1+h)}}h =\frac{h^{3/2}/12+o(h^{3/2})}h =\frac{\sqrt h}{12}+o(\sqrt h)\longrightarrow0.\)
Final result
\[0\]Exercise 6
Evaluate the following limit:
\[\lim_{x \to 0}\frac{\sin (2x) e^{-x}-\log(1+2x)}{x^{3}}.\]Solution.
Expand both factors and the logarithm to third order: \(\sin(2x)=2x-\frac43x^3+o(x^3).\) \(e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+o(x^3).\) \(\log(1+2x)=2x-2x^2+\frac83x^3+o(x^3).\) \(\sin(2x)e^{-x}=2x-2x^2+x^3-\frac43x^3+o(x^3).\) Subtracting cancels the first two orders: \(\frac{\sin(2x)e^{-x}-\log(1+2x)}{x^3} =\frac{-3x^3+o(x^3)}{x^3}\longrightarrow-3.\)
Final result
\[-3\]Exercise 7
Evaluate the following limit:
\[\lim_{x \to 0}\frac{e^{x-x^{2}}-\log(1+x)-1}{x-\sin x}.\]Solution.
Expand the exponential at the composite argument x − x²: \(e^{x-x^2}=1+x-x^2+\frac{(x-x^2)^2}{2}+\frac{(x-x^2)^3}{6}+o(x^3).\) \(e^{x-x^2}=1+x-\frac{x^2}{2}-\frac56x^3+o(x^3).\) \(\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3).\) \(e^{x-x^2}-\log(1+x)-1=-\frac76x^3+o(x^3).\) \(x-\sin x=\frac{x^3}{6}+o(x^3).\) \(\frac{-7x^3/6+o(x^3)}{x^3/6+o(x^3)}\longrightarrow-7.\)
Final result
\[-7\]Exercise 8
Evaluate the following limit:
\[\lim_{x \to 0}\frac{5^{1+\tan^{2}x}-5}{1-\cos x}.\]Solution.
Factor the constant exponential and use the tangent expansion: \(5^{1+\tan^2x}-5=5\left(e^{(\log5)\tan^2x}-1\right).\) \(e^{(\log5)\tan^2x}-1=(\log5)\tan^2x+o(\tan^2x).\) \(\tan^2x=x^2+o(x^2),\qquad1-\cos x=\frac{x^2}{2}+o(x^2).\) \(\frac{5(\log5)x^2+o(x^2)}{x^2/2+o(x^2)}\longrightarrow10\log5.\)
Final result
\[10\log5\]Exercise 9
Evaluate the following limit:
\[\lim_{x \to 0}\frac{x^{2}-\sin^{2} x}{x^{3}(e^{x}-\cos x)}.\]Solution.
Square the sine expansion before subtracting: \(\sin^2x=\left(x-\frac{x^3}{6}+o(x^3)\right)^2 =x^2-\frac{x^4}{3}+o(x^4).\) \(e^x-\cos x =1+x+\frac{x^2}{2}+o(x^2)-1+\frac{x^2}{2}+o(x^2) =x+x^2+o(x^2).\) \(\frac{x^2-\sin^2x}{x^3(e^x-\cos x)} =\frac{x^4/3+o(x^4)}{x^4+x^5+o(x^5)}\longrightarrow\frac13.\)
Final result
\[\frac13\]Exercise 10
Evaluate the following limit:
\[\lim_{x \to 1}\biggl(\frac{x}{x-1}-\frac{1}{\log x}\biggl).\]Solution.
Put y = x − 1, so x = 1 + y: \(\frac{x}{x-1}-\frac1{\log x} =\frac{(1+y)\log(1+y)-y}{y\log(1+y)}.\) \(\log(1+y)=y-\frac{y^2}{2}+o(y^2).\) \((1+y)\log(1+y)-y =y+y^2-\frac{y^2}{2}+o(y^2)-y =\frac{y^2}{2}+o(y^2).\) \(\frac{y^2/2+o(y^2)}{y^2(1-y/2+o(y))}\longrightarrow\frac12.\)
Final result
\[\frac12\]Exercise 11 — Expansion around x = 1
\[\lim_{x\to1}\frac{\log(1+\pi-4\arctan x)}{\sin(1-x)}.\]Solution.
First expand the logarithm and sine, whose arguments tend to zero:
\[\frac{\log(1+\pi-4\arctan x)}{\sin(1-x)} =\frac{\pi-4\arctan x+o(\pi-4\arctan x)}{(1-x)+o(1-x)}.\]Expand arctan at x = 1:
\[\arctan x=\frac\pi4+\frac{x-1}{2}+o(x-1).\]Substituting gives:
\[\frac{2(1-x)+o(1-x)+o(1-x)}{(1-x)+o(1-x)} =\frac{2+o(1)}{1+o(1)}\longrightarrow2.\]Final Result
\[2\]Exercise 12 — Fourth-order cancellation
\[\lim_{x\to0}\frac{x\arcsin x-x^2}{\sqrt{1+x^4}-\cos(x^2)}.\]Solution.
Use the expansions:
\[\arcsin x=x+\frac{x^3}{6}+o(x^3).\] \[\sqrt{1+x^4}=1+\frac{x^4}{2}+o(x^4), \qquad \cos(x^2)=1-\frac{x^4}{2}+o(x^4).\]Multiply the arcsine expansion by x and cancel the quadratic terms:
\[x\left(x+\frac{x^3}{6}+o(x^3)\right)-x^2 =\frac{x^4}{6}+o(x^4).\]The denominator is:
\[1+\frac{x^4}{2}+o(x^4)-1+\frac{x^4}{2}+o(x^4) =x^4+o(x^4).\]Therefore:
\[\frac{x^4/6+o(x^4)}{x^4+o(x^4)}\longrightarrow\frac16.\]Final Result
\[\frac16\]Exercise 13 — A product with fourth-order cancellation
\[\lim_{x\to0}\frac{\sin(3x)\log(1+2x)-6x^2+6x^3}{x^4}.\]Solution.
Expand each factor to the required order:
\[\sin(3x)=3x-\frac{27}{6}x^3+o(x^3).\] \[\log(1+2x)=2x-2x^2+\frac83x^3-4x^4+o(x^4).\]Multiplying and applying the little-o rules gives:
\[\sin(3x)\log(1+2x) =6x^2-6x^3+8x^4-9x^4+o(x^4).\]Thus the quotient becomes:
\[\frac{6x^2-6x^3+8x^4-9x^4+o(x^4)-6x^2+6x^3}{x^4} =\frac{-x^4+o(x^4)}{x^4}\longrightarrow-1.\]Final Result
\[-1\]Exercise 14 — A fifth root and the binomial expansion
\[\lim_{x\to0}\frac{\sqrt[5]{1-5x^2+x^4}-1+x^2}{x^4}.\]Solution.
Use the binomial expansion with exponent 1/5 and argument −5x²+x⁴:
\[(1-5x^2+x^4)^{1/5} =1+\frac{-5x^2+x^4}{5} -\frac2{25}(-5x^2+x^4)^2 +o((-5x^2+x^4)^2).\]The square contributes 25x⁴ at the required order. Hence:
\[\frac{1-x^2+x^4/5-2x^4-1+x^2+o(x^4)}{x^4} =-\frac95+o(1).\]Final Result
\[-\frac95\]Exercise 15 — Matching eighth-order terms
\[\lim_{x\to0}\frac{x^5(e^{2x^3}-1)}{\sqrt{1+x^8}-\sqrt[3]{1+x^8}}.\]Solution.
Apply the exponential and binomial expansions:
\[e^{2x^3}=1+2x^3+o(x^3).\] \[\sqrt{1+x^8}=1+\frac{x^8}{2}+o(x^8), \qquad \sqrt[3]{1+x^8}=1+\frac{x^8}{3}+o(x^8).\]Then:
\[\frac{x^5(1+2x^3+o(x^3)-1)}{1+x^8/2+o(x^8)-1-x^8/3+o(x^8)} =\frac{2x^8+o(x^8)}{x^8/6+o(x^8)}\longrightarrow12.\]Final Result
\[12\]