Notable Limits in Calculus: 14 Solved Examples
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.
Notable and Remarkable Limits in Calculus
Fundamental Limit Formulas
The following formulas use natural logarithms and trigonometric arguments measured in radians:
\[\lim_{x\to 0} \frac{\sin x}{x} = 1, \quad \lim_{x\to \infty} \frac{\sin x}{x} = 0, \quad \lim_{x\to 0} \frac{\log(1+x)}{x} = 1,\] \[\lim_{x\to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x\to \pm\infty} \left(1 + \frac{1}{x}\right)^x = e, \quad \lim_{x\to 0} (1+x)^{1/x} = e,\] \[\lim_{x\to 0} \frac{e^x - 1}{x} = 1, \quad \lim_{x\to 0} \frac{\tan x}{x} = 1, \quad \lim_{x\to 0} \frac{\arcsin x}{x} = 1, \quad \lim_{x\to 0} \frac{\arctan x}{x} = 1.\]Key Theorems
Theorem: Non-Existence via Sequences
Let c be an accumulation point of the domain. If two sequences aₙ → c and bₙ → c lie in that domain, satisfy aₙ ≠ c and bₙ ≠ c for every n, and have existing but distinct image limits:
\[\lim_{n \to \infty} f(a_n) \ne \lim_{n \to \infty} f(b_n),\]then
\[\lim_{x \to c} f(x)\]does not exist.
Theorem: Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) near x = c, and:
\[\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L,\]then:
\[\lim_{x \to c} g(x) = L.\]Step-by-Step Solutions
Exercise 1
\[\lim_{x \to +\infty} \left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x + 1} \right)\]Solution.
We multiply and divide by the conjugate expression:
\[\begin{aligned} \lim_{x \to +\infty} &\left( \sqrt{x^2 + x + 1} - \sqrt{x^2 - x + 1} \right) \\ &= \lim_{x \to +\infty} \frac{(x^2 + x + 1) - (x^2 - x + 1)} {\sqrt{x^2 + x + 1} + \sqrt{x^2 - x + 1}} \\ &= \lim_{x \to +\infty} \frac{2x} {\sqrt{x^2 + x + 1} + \sqrt{x^2 - x + 1}} \\ &= \lim_{x \to +\infty} \frac{2x} {x \left( \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} \right)} \\ &= \frac{2}{1+1}. \end{aligned}\]Final result
\[1\]Exercise 2
\[\lim_{x \to \infty} x \cdot \log\left( \frac{x + 4}{x + 5} \right)\]Solution.
We rewrite the logarithmic expression:
\[\lim_{x \to \infty} x \cdot \log\left( 1 - \frac{1}{x + 5} \right).\]Let:
\[t = -\frac{1}{x+5}.\]Then:
\[x = \frac{-1-5t}{t}.\]Therefore:
\[\lim_{t \to 0} \frac{-1-5t}{t}\log(1+t).\]Using the fundamental limit:
\[\lim_{t\to0}\frac{\log(1+t)}{t}=1,\]we obtain:
\[\lim_{t \to 0} (-1-5t)\frac{\log(1+t)}{t} = -1.\]Final result
\[-1\]Exercise 3
\[\lim_{x \to +\infty} \left( \frac{2x + 9}{2x + 1} \right)^x\]Solution.
We write:
\[\left( \frac{2x + 9}{2x + 1} \right)^x = \left( 1 + \frac{8}{2x + 1} \right)^x.\]Let:
\[t = \frac{8}{2x+1}.\]Then:
\[x = \frac{4}{t}-\frac{1}{2}.\]Therefore:
\[(1+t)^{\frac{4}{t}-\frac{1}{2}} = \frac{(1+t)^{4/t}}{(1+t)^{1/2}}.\]Using:
\[\lim_{t\to0}(1+t)^{1/t}=e,\]we obtain:
\[\frac{e^4}{1}=e^4.\]Final result
\[e^4\]Exercise 4
\[\lim_{x \to +\infty} x \cdot \log\left( \frac{x^2 + 1}{x^2 + x} \right)\]Solution.
We simplify the ratio:
\[\frac{x^2 + 1}{x^2 + x} = \frac{1 + \frac{1}{x^2}} {1 + \frac{1}{x}}.\]Let:
\[t = \frac{1}{x}.\]Then:
\[x = \frac{1}{t}.\]Therefore:
\[\lim_{t \to 0} \frac{1}{t} \log\left( \frac{1+t^2}{1+t} \right).\]Using the logarithm of a quotient:
\[\lim_{t \to 0} \left( \frac{\log(1+t^2)}{t} - \frac{\log(1+t)}{t} \right).\]For the first term:
\[\frac{\log(1+t^2)}{t} = t\frac{\log(1+t^2)}{t^2} \to 0.\]For the second:
\[\frac{\log(1+t)}{t}\to1.\]Hence:
\[0-1=-1.\]Final result
\[-1\]Exercise 5
\[\lim_{x \to +\infty} \frac{\log(x^3 + 1)}{x}\]Solution.
We use the identity:
\[\log(x^3 + 1) = \log\left( x^3 \left( 1 + \frac{1}{x^3} \right) \right).\]Therefore:
\[\log(x^3+1) = 3\log x + \log\left( 1+\frac{1}{x^3} \right).\]Hence:
\[\frac{\log(x^3+1)}{x} = \frac{3\log x}{x} + \frac{ \log\left(1+\frac{1}{x^3}\right) }{x}.\]Both terms tend to zero:
\[\frac{3\log x}{x}\to0,\]and
\[\frac{ \log\left(1+\frac{1}{x^3}\right) }{x}\to0.\]Final result
\[0\]Exercise 6
\[\lim_{x \to +\infty} \frac{\sin x - x} {\cos x + \sqrt{1 + x^2}}\]Solution.
We divide numerator and denominator by x:
\[\frac{ \frac{\sin x}{x}-1 }{ \frac{\cos x}{x} + \sqrt{1+\frac{1}{x^2}} }.\]As x → +∞:
\[\frac{\sin x}{x}\to0, \qquad \frac{\cos x}{x}\to0,\]and:
\[\sqrt{1+\frac{1}{x^2}}\to1.\]Therefore:
\[\frac{0-1}{0+1}=-1.\]Final result
\[-1\]Exercise 7
\[\lim_{x \to +\infty} \sin x \left[ \log(\sqrt{x}+1) - \log(\sqrt{x+1}) \right]\]Solution.
We use the logarithm of a quotient:
\[\log(\sqrt{x}+1) - \log(\sqrt{x+1}) = \log\left( \frac{\sqrt{x}+1}{\sqrt{x+1}} \right).\]Now:
\[\frac{\sqrt{x}+1}{\sqrt{x+1}} = \frac{ \sqrt{x} \left( 1+\frac{1}{\sqrt{x}} \right) }{ \sqrt{x} \sqrt{ 1+\frac{1}{x} } }.\]Thus:
\[\frac{\sqrt{x}+1}{\sqrt{x+1}} = \frac{ 1+\frac{1}{\sqrt{x}} }{ \sqrt{ 1+\frac{1}{x} } }.\]As x → +∞, the ratio tends to 1, so:
\[\log\left( \frac{\sqrt{x}+1}{\sqrt{x+1}} \right) \to0.\]Since sin x is bounded, the product tends to zero.
Final result
\[0\]Exercise 8
\[\lim_{x \to +\infty} \left( \frac{x+3}{x-1} \right)^{x+1}\]Solution.
We rewrite:
\[\frac{x+3}{x-1} = 1+\frac{4}{x-1}.\]Therefore:
\[\left( 1+\frac{4}{x-1} \right)^{x+1}.\]Let:
\[y=\frac{4}{x-1}.\]Then:
\[x=1+\frac{4}{y},\]and hence:
\[x+1=2+\frac{4}{y}.\]Therefore:
\[(1+y)^{2+\frac{4}{y}} = (1+y)^2 \left( (1+y)^{1/y} \right)^4.\]As y → 0:
\[(1+y)^2\to1,\]while:
\[(1+y)^{1/y}\to e.\]Hence:
\[1\cdot e^4=e^4.\]Final result
\[e^4\]Exercise 9
\[\lim_{x \to 0^+} x^{\frac{1}{\log(3x)}}\]Solution.
Let:
\[y=\frac{1}{\log(3x)}.\]As x → 0⁺, we have:
\[\log(3x)\to-\infty,\]so:
\[y\to0^-.\]From the definition of y:
\[\log(3x)=\frac{1}{y}.\]Therefore:
\[3x=e^{1/y},\]and:
\[x=\frac{e^{1/y}}{3}.\]The original expression becomes:
\[x^y = \left( \frac{e^{1/y}}{3} \right)^y.\]Hence:
\[x^y = \frac{ e^{(1/y)y} }{ 3^y }.\]Therefore:
\[x^y = \frac{e}{3^y}.\]As y → 0⁻:
\[3^y\to1.\]Thus:
\[\frac{e}{3^y}\to e.\]Final result
\[e\]Exercise 10
\[\lim_{x \to 0} \frac{e^x-e^{-x}}{x}\]Solution.
We split the expression:
\[\frac{e^x-e^{-x}}{x} = \frac{e^x-1}{x} + \frac{1-e^{-x}}{x}.\]The second term can be rewritten as:
\[\frac{1-e^{-x}}{x} = \frac{e^{-x}-1}{-x}.\]Therefore:
\[\frac{e^x-e^{-x}}{x} = \frac{e^x-1}{x} + \frac{e^{-x}-1}{-x}.\]Using the fundamental limits:
\[\lim_{x \to 0} \frac{e^x-1}{x} = 1,\]and:
\[\lim_{x \to 0} \frac{e^{-x}-1}{-x} = 1,\]we obtain:
\[1+1=2.\]Final result
\[2\]Further Notable Limits
These additional exercises and their solutions are translated from Eserciziario 2.1 by Antonino De Martino and Luana Manfredini. No new exercise statements have been introduced.
Exercise 11 — Rationalization and a cosine limit
\[\lim_{x\to0}\frac{\sqrt{1+x+x^2}-\sqrt{1+x}}{1-\cos(2x)}.\]Solution.
Multiply numerator and denominator by the conjugate:
\[\frac{(\sqrt{1+x+x^2}-\sqrt{1+x})(\sqrt{1+x+x^2}+\sqrt{1+x})} {(1-\cos(2x))(\sqrt{1+x+x^2}+\sqrt{1+x})}\] \[=\frac{1+x+x^2-1-x}{(1-\cos(2x))(\sqrt{1+x+x^2}+\sqrt{1+x})} =\frac{x^2}{(1-\cos(2x))(\sqrt{1+x+x^2}+\sqrt{1+x})}.\]Normalize the cosine term using its argument 2x:
\[\frac{x^2/(4x^2)}{[(1-\cos(2x))/(4x^2)](\sqrt{1+x+x^2}+\sqrt{1+x})} \longrightarrow\frac{1/4}{(1/2)\cdot2}.\]Final Result
\[\frac14\]Exercise 12 — A logarithmic limit at x = 1
\[\lim_{x\to1}\frac{\log x}{x^3-1}.\]Solution.
Set y = x³−1, so that x is the cube root of 1+y. The logarithm becomes:
\[\log x=\frac13\log(1+y).\]The notable logarithmic limit gives:
\[\lim_{y\to0}\frac{(1/3)\log(1+y)}y=\frac13.\]Final Result
\[\frac13\]Exercise 13 — A power with a parameter
\[\lim_{x\to0}\frac{(1+x)^\alpha-1}{x},\qquad\alpha>0.\]Solution.
Multiply and divide by log(1+x):
\[\frac{(1+x)^\alpha-1}{x} =\frac{(1+x)^\alpha-1}{\log(1+x)}\frac{\log(1+x)}x.\]Set y = (1+x)ᵅ−1. Then:
\[(1+x)^\alpha=1+y, \qquad\log(1+x)=\frac{\log(1+y)}\alpha.\]Therefore the limit is:
\[\lim_{y\to0}\alpha\frac y{\log(1+y)} \cdot\lim_{x\to0}\frac{\log(1+x)}x=\alpha.\]Final Result
\[\alpha\]Exercise 14 — Two parameters and three cases
\[\lim_{x\to0}\frac{\sin(|x|^\alpha)}{|x|^\beta},\qquad\alpha,\beta>0.\]Solution.
Multiply and divide by ∣x∣ᵅ:
\[\frac{\sin(|x|^\alpha)}{|x|^\beta} =\frac{\sin(|x|^\alpha)}{|x|^\alpha}|x|^{\alpha-\beta}.\]The first factor tends to 1 by the sine notable limit. The second factor tends to zero for α > β, equals 1 for α = β, and tends to +∞ for α < β. These exhaust the three possible parameter comparisons.
Final Result
\[\begin{cases} 0,&\alpha>\beta,\\ 1,&\alpha=\beta,\\ +\infty,&\alpha<\beta. \end{cases}\]Continue Exploring Limits
These examples show how algebraic transformations, fundamental limits, logarithms, exponentials, bounded functions, and substitutions can be combined to evaluate apparently different limiting forms.