The following practice problems progressively apply the definitions and properties of infinite quantities, infinitesimals, orders, asymptotic equivalence, and little-o notation.
In each problem the limiting process is stated explicitly, since a function may tend to infinity, tend to zero, or exhibit neither behaviour, depending on the limit under consideration.
Exercise 1 — level ★☆☆☆☆
Determine whether the function
\[ f(x)=x^3 \]
is infinitesimal or tends to infinity in each of the limiting processes \(x\to 0\) and \(x\to+\infty\).
Answer
The function \(x^3\) is infinitesimal as \(x\to 0\) and tends to \(+\infty\) as \(x\to+\infty\).
Solution
A function is said to be infinitesimal with respect to a given limiting process when its limit equals zero.
Consider first \(x\to 0\). Since \(x^3\) is continuous, the limit may be evaluated by direct substitution:
\[ \lim_{x\to 0}x^3=0^3=0. \]
Hence \(x^3\) is infinitesimal as \(x\to 0\).
Now consider \(x\to+\infty\). As \(x\) takes larger and larger positive values, so does \(x^3\). Thus
\[ \lim_{x\to+\infty}x^3=+\infty. \]
Consequently, \(x^3\) tends to \(+\infty\) as \(x\to+\infty\).
This problem illustrates that whether a function tends to infinity or is infinitesimal depends on the limiting process under consideration.
Exercise 2 — level ★☆☆☆☆
Study the behaviour of the function
\[ f(x)=\frac{1}{x^2} \]
as \(x\to 0\) and as \(x\to+\infty\).
Answer
The function \(\displaystyle\frac{1}{x^2}\) tends to \(+\infty\) as \(x\to 0\) and is infinitesimal as \(x\to+\infty\).
Solution
Consider first the limit as \(x\to 0\).
As \(x\) approaches zero, \(x^2\) remains positive and tends to zero:
\[ x^2\to 0^+. \]
The reciprocal of a positive quantity tending to zero takes arbitrarily large positive values. Hence
\[ \lim_{x\to 0}\frac{1}{x^2}=+\infty. \]
The function therefore tends to \(+\infty\) as \(x\to 0\).
Now consider \(x\to+\infty\). In this case
\[ x^2\to+\infty. \]
The reciprocal of a positive quantity tending to \(+\infty\) tends to zero. It follows that
\[ \lim_{x\to+\infty}\frac{1}{x^2}=0. \]
Hence \(\displaystyle\frac{1}{x^2}\) is infinitesimal as \(x\to+\infty\).
Exercise 3 — level ★★☆☆☆
Compare the rates of growth of the functions
\[ f(x)=x^2 \qquad\text{and}\qquad g(x)=x^5 \]
as \(x\to+\infty\).
Answer
The function \(x^2\) is of lower order than \(x^5\).
Solution
Both functions tend to \(+\infty\), since
\[ \lim_{x\to+\infty}x^2=+\infty \qquad\text{and}\qquad \lim_{x\to+\infty}x^5=+\infty. \]
To compare the rate at which they grow, we evaluate the limit of the ratio of \(f(x)\) to \(g(x)\):
\[ \lim_{x\to+\infty}\frac{f(x)}{g(x)} = \lim_{x\to+\infty}\frac{x^2}{x^5}. \]
Using the properties of powers, we simplify:
\[ \frac{x^2}{x^5}=\frac{1}{x^3}. \]
Since \(x^3\to+\infty\), its reciprocal tends to zero:
\[ \lim_{x\to+\infty}\frac{1}{x^3}=0. \]
We have thus obtained
\[ \lim_{x\to+\infty}\frac{x^2}{x^5}=0. \]
When the ratio of two functions tending to infinity tends to zero, the function in the numerator is of lower order than the function in the denominator.
Hence \(x^2\) is of lower order than \(x^5\).
Exercise 4 — level ★★☆☆☆
Compare the infinitesimals
\[ f(x)=\frac{1}{x^4} \qquad\text{and}\qquad g(x)=\frac{1}{x} \]
as \(x\to+\infty\).
Answer
The function \(\displaystyle\frac{1}{x^4}\) is an infinitesimal of higher order than \(\displaystyle\frac{1}{x}\).
Solution
Both functions are infinitesimal as \(x\to+\infty\), since
\[ \lim_{x\to+\infty}\frac{1}{x^4}=0 \qquad\text{and}\qquad \lim_{x\to+\infty}\frac{1}{x}=0. \]
To compare two infinitesimals we consider the limit of their ratio:
\[ \lim_{x\to+\infty} \frac{\displaystyle\frac{1}{x^4}} {\displaystyle\frac{1}{x}}. \]
Dividing by a fraction is equivalent to multiplying by its reciprocal. Hence
\[ \frac{\displaystyle\frac{1}{x^4}} {\displaystyle\frac{1}{x}} = \frac{1}{x^4}\cdot x = \frac{1}{x^3}. \]
We evaluate the limit:
\[ \lim_{x\to+\infty}\frac{1}{x^3}=0. \]
When the ratio of two infinitesimals tends to zero, the infinitesimal in the numerator tends to zero more rapidly and is therefore of higher order.
Consequently, \(\displaystyle\frac{1}{x^4}\) is an infinitesimal of higher order than \(\displaystyle\frac{1}{x}\).
Exercise 5 — level ★★☆☆☆
Determine whether the functions
\[ f(x)=3x^2+1 \qquad\text{and}\qquad g(x)=x^2 \]
are of the same order and whether they are asymptotically equivalent as \(x\to+\infty\).
Answer
The two functions are of the same order, but they are not asymptotically equivalent.
Solution
As \(x\to+\infty\), both functions tend to \(+\infty\). We may therefore compare them by means of their ratio:
\[ \lim_{x\to+\infty}\frac{3x^2+1}{x^2}. \]
We divide each term of the numerator separately by \(x^2\):
\[ \frac{3x^2+1}{x^2} = \frac{3x^2}{x^2}+\frac{1}{x^2} = 3+\frac{1}{x^2}. \]
Since
\[ \lim_{x\to+\infty}\frac{1}{x^2}=0, \]
we obtain
\[ \lim_{x\to+\infty}\frac{3x^2+1}{x^2}=3. \]
The ratio tends to a finite, nonzero constant. The two functions are therefore of the same order.
To be asymptotically equivalent, however, the ratio would need to tend to exactly \(1\). Since the limit equals \(3\), we cannot write
\[ 3x^2+1\sim x^2. \]
We may instead observe that
\[ 3x^2+1\sim 3x^2, \]
because
\[ \lim_{x\to+\infty}\frac{3x^2+1}{3x^2}=1. \]
Exercise 6 — level ★★☆☆☆
Verify that
\[ x^2+x\sim x^2 \qquad\text{as }x\to+\infty. \]
Answer
The equivalence holds:
\[ x^2+x\sim x^2 \qquad\text{as }x\to+\infty. \]
Solution
Two functions are asymptotically equivalent when the ratio of the first to the second tends to \(1\).
We must therefore evaluate
\[ \lim_{x\to+\infty}\frac{x^2+x}{x^2}. \]
We divide both terms of the numerator by \(x^2\):
\[ \frac{x^2+x}{x^2} = \frac{x^2}{x^2}+\frac{x}{x^2} = 1+\frac{1}{x}. \]
Since
\[ \lim_{x\to+\infty}\frac{1}{x}=0, \]
it follows that
\[ \lim_{x\to+\infty} \left(1+\frac{1}{x}\right)=1. \]
Hence
\[ \lim_{x\to+\infty}\frac{x^2+x}{x^2}=1, \]
and therefore
\[ x^2+x\sim x^2. \]
The term \(x\) is negligible compared with \(x^2\) in the leading behaviour of the function as \(x\to+\infty\).
Exercise 7 — level ★★★☆☆
Determine the order of the function
\[ f(x)=x^6 \]
with respect to the function
\[ g(x)=x^2 \]
as \(x\to+\infty\).
Answer
The function \(x^6\) is of order \(3\) with respect to \(x^2\).
Solution
A function \(f\) tending to infinity is said to be of order \(\alpha>0\) with respect to a function \(g\) tending to infinity if
\[ \lim_{x\to+\infty}\frac{f(x)}{[g(x)]^\alpha}=L, \qquad 0<|L|<+\infty. \]
Here
\[ f(x)=x^6 \qquad\text{and}\qquad g(x)=x^2. \]
We seek a number \(\alpha>0\) such that
\[ [g(x)]^\alpha=(x^2)^\alpha=x^{2\alpha} \]
has the same power of \(x\) as \(f(x)=x^6\).
We must therefore impose
\[ 2\alpha=6. \]
Dividing both sides by \(2\), we obtain
\[ \alpha=3. \]
Let us verify this against the definition:
\[ \lim_{x\to+\infty} \frac{x^6}{(x^2)^3} = \lim_{x\to+\infty} \frac{x^6}{x^6} = \lim_{x\to+\infty}1 = 1. \]
The limit is finite and nonzero. Hence \(x^6\) is of order \(3\) with respect to \(x^2\).
Exercise 8 — level ★★★☆☆
Determine the order of the infinitesimal
\[ f(x)=x^5 \]
with respect to the infinitesimal
\[ g(x)=x \]
as \(x\to 0^+\).
Answer
The function \(x^5\) is an infinitesimal of order \(5\) with respect to \(x\).
Solution
As \(x\to 0^+\), both \(x^5\) and \(x\) tend to zero. They are therefore two infinitesimals with respect to the same limiting process.
We seek \(\alpha>0\) such that
\[ \lim_{x\to 0^+}\frac{x^5}{x^\alpha} \]
is a finite, nonzero constant.
Using the properties of powers, for \(x>0\) we have
\[ \frac{x^5}{x^\alpha}=x^{5-\alpha}. \]
For the ratio to be identically equal to \(1\), we must choose
\[ 5-\alpha=0. \]
It follows that
\[ \alpha=5. \]
Let us verify:
\[ \lim_{x\to 0^+}\frac{x^5}{x^5} = \lim_{x\to 0^+}1 = 1. \]
Since the limit is finite and nonzero, \(x^5\) is an infinitesimal of order \(5\) with respect to \(x\).
Exercise 9 — level ★★★☆☆
Determine whether the functions
\[ f(x)=x(2+\sin x) \qquad\text{and}\qquad g(x)=x \]
are comparable under the classification based on the limit of their ratio as \(x\to+\infty\).
Answer
Both functions tend to \(+\infty\), but they are not comparable under the classification based on the limit of their ratio, because that limit does not exist.
Solution
For every real number \(x\),
\[ -1\leq\sin x\leq 1. \]
Adding \(2\) to all sides gives
\[ 1\leq 2+\sin x\leq 3. \]
For \(x>0\), multiplying by \(x\) yields
\[ x\leq x(2+\sin x)\leq 3x. \]
Since \(x\to+\infty\), it follows that \(x(2+\sin x)\to+\infty\) as well. Hence both \(f\) and \(g\) tend to \(+\infty\).
We now compute their ratio:
\[ \frac{f(x)}{g(x)} = \frac{x(2+\sin x)}{x} = 2+\sin x. \]
The function \(\sin x\) continues to oscillate between \(-1\) and \(1\) as \(x\to+\infty\). Consequently, \(2+\sin x\) continues to oscillate between \(1\) and \(3\) and admits no limit.
The ratio therefore neither tends to zero nor to a finite nonzero constant, and its absolute value does not tend to \(+\infty\).
The two functions are consequently not comparable under this classification.
Exercise 10 — level ★★★☆☆
Study the relationship between the infinitesimal
\[ f(x)=x \]
and its reciprocal as \(x\to 0\), distinguishing the right-hand limit, the left-hand limit, and the two-sided limit.
Answer
We have
\[ \lim_{x\to 0^+}\frac{1}{x}=+\infty, \qquad \lim_{x\to 0^-}\frac{1}{x}=-\infty. \]
The two-sided limit of \(\displaystyle\frac{1}{x}\) as \(x\to 0\) does not exist, whereas
\[ \lim_{x\to 0}\left|\frac{1}{x}\right|=+\infty. \]
Solution
The function \(f(x)=x\) is infinitesimal as \(x\to 0\), since
\[ \lim_{x\to 0}x=0. \]
However, \(x\) does not have a definite sign in a full neighbourhood of zero: it is positive to the right of zero and negative to the left.
Consider first \(x\to 0^+\). In this case \(x\) takes smaller and smaller positive values:
\[ x\to 0^+. \]
The reciprocal of a positive quantity tending to zero tends to \(+\infty\). Hence
\[ \lim_{x\to 0^+}\frac{1}{x}=+\infty. \]
Now consider \(x\to 0^-\). In this case \(x\) takes negative values whose absolute value becomes smaller and smaller:
\[ x\to 0^-. \]
The reciprocal remains negative and grows arbitrarily large in absolute value. Consequently
\[ \lim_{x\to 0^-}\frac{1}{x}=-\infty. \]
The two one-sided limits differ. The two-sided limit
\[ \lim_{x\to 0}\frac{1}{x} \]
therefore does not exist, even in the extended real sense.
The absolute value of the reciprocal, on the other hand, is
\[ \left|\frac{1}{x}\right|=\frac{1}{|x|}. \]
Since \(|x|\to 0^+\), we have
\[ \lim_{x\to 0}\frac{1}{|x|}=+\infty. \]
Hence
\[ \lim_{x\to 0}\left|\frac{1}{x}\right|=+\infty. \]
Exercise 11 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{\sin(3x)}{x}. \]
Answer
\[ \lim_{x\to 0}\frac{\sin(3x)}{x}=3. \]
Solution
For \(u\to 0\) the fundamental equivalence
\[ \sin u\sim u \]
holds. Here the argument of the sine is
\[ u=3x. \]
As \(x\to 0\), we also have \(3x\to 0\). We may therefore apply the equivalence:
\[ \sin(3x)\sim 3x. \]
Since \(\sin(3x)\) appears as a factor in the numerator of a quotient, we may replace it with its equivalent:
\[ \lim_{x\to 0}\frac{\sin(3x)}{x} = \lim_{x\to 0}\frac{3x}{x}. \]
For \(x\neq 0\), we cancel \(x\):
\[ \frac{3x}{x}=3. \]
Hence
\[ \lim_{x\to 0}\frac{\sin(3x)}{x} = \lim_{x\to 0}3 = 3. \]
Exercise 12 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{1-\cos(2x)}{x^2}. \]
Answer
\[ \lim_{x\to 0}\frac{1-\cos(2x)}{x^2}=2. \]
Solution
For \(u\to 0\) the equivalence
\[ 1-\cos u\sim\frac{u^2}{2} \]
holds. Set
\[ u=2x. \]
As \(x\to 0\), we also have \(u=2x\to 0\). We may therefore write
\[ 1-\cos(2x)\sim\frac{(2x)^2}{2}. \]
We compute the square:
\[ (2x)^2=4x^2. \]
Consequently
\[ \frac{(2x)^2}{2} = \frac{4x^2}{2} = 2x^2. \]
Hence
\[ 1-\cos(2x)\sim 2x^2. \]
Substituting the equivalent into the quotient, we obtain
\[ \lim_{x\to 0}\frac{1-\cos(2x)}{x^2} = \lim_{x\to 0}\frac{2x^2}{x^2}. \]
For \(x\neq 0\),
\[ \frac{2x^2}{x^2}=2. \]
The limit therefore equals
\[ 2. \]
Exercise 13 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{e^{2x}-1}{\tan(5x)}. \]
Answer
\[ \lim_{x\to 0}\frac{e^{2x}-1}{\tan(5x)}=\frac{2}{5}. \]
Solution
For \(u\to 0\) the equivalences
\[ e^u-1\sim u \qquad\text{and}\qquad \tan u\sim u \]
hold. In the numerator we set \(u=2x\). Since \(2x\to 0\), we obtain
\[ e^{2x}-1\sim 2x. \]
In the denominator we set \(u=5x\). Since \(5x\to 0\), we obtain
\[ \tan(5x)\sim 5x. \]
These two expressions appear respectively in the numerator and denominator of a quotient. We may therefore use their equivalents:
\[ \lim_{x\to 0}\frac{e^{2x}-1}{\tan(5x)} = \lim_{x\to 0}\frac{2x}{5x}. \]
For \(x\neq 0\), we cancel \(x\):
\[ \frac{2x}{5x}=\frac{2}{5}. \]
Hence
\[ \lim_{x\to 0}\frac{e^{2x}-1}{\tan(5x)} = \frac{2}{5}. \]
Exercise 14 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{\ln(1+4x)}{\arcsin(3x)}. \]
Answer
\[ \lim_{x\to 0}\frac{\ln(1+4x)}{\arcsin(3x)}=\frac{4}{3}. \]
Solution
For \(u\to 0\) the equivalences
\[ \ln(1+u)\sim u \qquad\text{and}\qquad \arcsin u\sim u \]
hold. In the numerator, the increment appearing inside the logarithm is \(4x\). As \(x\to 0\), we also have \(4x\to 0\). Hence
\[ \ln(1+4x)\sim 4x. \]
In the denominator, the argument of the arcsine is \(3x\). Since \(3x\to 0\), we have
\[ \arcsin(3x)\sim 3x. \]
Using the equivalents in the quotient, we obtain
\[ \lim_{x\to 0}\frac{\ln(1+4x)}{\arcsin(3x)} = \lim_{x\to 0}\frac{4x}{3x}. \]
For \(x\neq 0\), we cancel the factor \(x\):
\[ \frac{4x}{3x}=\frac{4}{3}. \]
The required limit is therefore
\[ \frac{4}{3}. \]
Exercise 15 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{(1+x)^5-1}{e^{2x}-1}. \]
Answer
\[ \lim_{x\to 0}\frac{(1+x)^5-1}{e^{2x}-1}=\frac{5}{2}. \]
Solution
For \(x\to 0\) the equivalence
\[ (1+x)^\alpha-1\sim\alpha x, \qquad \alpha\neq 0 \]
holds. Here \(\alpha=5\). Hence
\[ (1+x)^5-1\sim 5x. \]
For the denominator we use the equivalence
\[ e^u-1\sim u \qquad\text{as }u\to 0. \]
Setting \(u=2x\), and since \(2x\to 0\), we obtain
\[ e^{2x}-1\sim 2x. \]
The two expressions occur in a quotient, so we may substitute their respective equivalents:
\[ \lim_{x\to 0}\frac{(1+x)^5-1}{e^{2x}-1} = \lim_{x\to 0}\frac{5x}{2x}. \]
For \(x\neq 0\), we cancel \(x\):
\[ \frac{5x}{2x}=\frac{5}{2}. \]
Hence
\[ \lim_{x\to 0}\frac{(1+x)^5-1}{e^{2x}-1} = \frac{5}{2}. \]
Exercise 16 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{\sin(2x)\tan(3x)}{x^2}. \]
Answer
\[ \lim_{x\to 0}\frac{\sin(2x)\tan(3x)}{x^2}=6. \]
Solution
The numerator is a product of two infinitesimals. We may apply the fundamental equivalences to each factor separately.
Since
\[ \sin u\sim u \qquad\text{as }u\to 0, \]
setting \(u=2x\) gives
\[ \sin(2x)\sim 2x. \]
Moreover
\[ \tan u\sim u \qquad\text{as }u\to 0. \]
Setting \(u=3x\), we obtain
\[ \tan(3x)\sim 3x. \]
The equivalents are used within a product and a quotient, so the substitution is legitimate:
\[ \lim_{x\to 0}\frac{\sin(2x)\tan(3x)}{x^2} = \lim_{x\to 0}\frac{(2x)(3x)}{x^2}. \]
We multiply the factors in the numerator:
\[ (2x)(3x)=6x^2. \]
Hence
\[ \lim_{x\to 0}\frac{6x^2}{x^2}. \]
For \(x\neq 0\),
\[ \frac{6x^2}{x^2}=6. \]
The required limit is therefore
\[ 6. \]
Exercise 17 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{1-\cos(4x)}{x\sin(2x)}. \]
Answer
\[ \lim_{x\to 0}\frac{1-\cos(4x)}{x\sin(2x)}=4. \]
Solution
For the numerator we use the equivalence
\[ 1-\cos u\sim\frac{u^2}{2} \qquad\text{as }u\to 0. \]
Setting \(u=4x\), we obtain
\[ 1-\cos(4x)\sim\frac{(4x)^2}{2}. \]
We compute:
\[ \frac{(4x)^2}{2} = \frac{16x^2}{2} = 8x^2. \]
Hence
\[ 1-\cos(4x)\sim 8x^2. \]
In the denominator we have the product \(x\sin(2x)\). For the sine,
\[ \sin(2x)\sim 2x. \]
Consequently
\[ x\sin(2x)\sim x\cdot 2x=2x^2. \]
We may now substitute the equivalents into the quotient:
\[ \lim_{x\to 0}\frac{1-\cos(4x)}{x\sin(2x)} = \lim_{x\to 0}\frac{8x^2}{2x^2}. \]
For \(x\neq 0\), we cancel \(x^2\):
\[ \frac{8x^2}{2x^2}=\frac{8}{2}=4. \]
The limit therefore equals
\[ 4. \]
Exercise 18 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\frac{(x^2+x)-x^2}{x} \]
and explain why it is incorrect to replace \(x^2+x\) directly with the equivalent \(x^2\) inside the difference.
Answer
\[ \lim_{x\to+\infty}\frac{(x^2+x)-x^2}{x}=1. \]
Substituting equivalents directly inside a difference is not legitimate here, because doing so would eliminate precisely the term that determines the value of the limit.
Solution
We know that
\[ x^2+x\sim x^2 \qquad\text{as }x\to+\infty, \]
because
\[ \lim_{x\to+\infty}\frac{x^2+x}{x^2}=1. \]
However, the expression \(x^2+x\) appears inside the difference
\[ (x^2+x)-x^2. \]
Equivalents cannot be substituted automatically inside sums and differences, because the leading terms may cancel.
We therefore compute the difference exactly:
\[ (x^2+x)-x^2=x^2+x-x^2=x. \]
The limit becomes
\[ \lim_{x\to+\infty}\frac{x}{x}. \]
For \(x\neq 0\),
\[ \frac{x}{x}=1. \]
Hence
\[ \lim_{x\to+\infty}\frac{(x^2+x)-x^2}{x}=1. \]
Had we substituted \(x^2+x\) with \(x^2\) directly inside the difference, we would have obtained
\[ x^2-x^2=0 \]
and hence, erroneously,
\[ \frac{0}{x}=0. \]
This substitution would have eliminated the term \(x\), which is negligible compared with \(x^2\) in the function \(x^2+x\), but which becomes the determining term once the \(x^2\) terms cancel.
Exercise 19 — level ★★★★☆
Using the standard hierarchies of growth, establish the following relations as \(x\to+\infty\):
\[ \ln(x)=o(\sqrt{x}), \qquad x^4=o(2^x), \]
and derive the corresponding relations between the reciprocals.
Answer
As \(x\to+\infty\),
\[ \ln(x)=o(\sqrt{x}), \qquad x^4=o(2^x), \]
and, passing to the reciprocals,
\[ \frac{1}{2^x}=o\left(\frac{1}{x^4}\right), \qquad \frac{1}{\sqrt{x}}=o\left(\frac{1}{\ln(x)}\right). \]
Solution
In the present limiting process, the notation
\[ f(x)=o(g(x)) \]
means that
\[ \lim_{x\to+\infty}\frac{f(x)}{g(x)}=0. \]
The standard hierarchy between logarithms and powers states that, for every \(\alpha>0\),
\[ \lim_{x\to+\infty}\frac{\ln(x)}{x^\alpha}=0. \]
Since
\[ \sqrt{x}=x^{1/2} \]
and \(\displaystyle\frac{1}{2}>0\), choosing \(\alpha=\displaystyle\frac{1}{2}\) gives
\[ \lim_{x\to+\infty}\frac{\ln(x)}{\sqrt{x}}=0. \]
By the definition of little-o, we may therefore write
\[ \ln(x)=o(\sqrt{x}). \]
The hierarchy between powers and exponentials further states that, for every \(\alpha>0\) and every \(a>1\),
\[ \lim_{x\to+\infty}\frac{x^\alpha}{a^x}=0. \]
Choosing \(\alpha=4\) and \(a=2\), we obtain
\[ \lim_{x\to+\infty}\frac{x^4}{2^x}=0. \]
Hence
\[ x^4=o(2^x). \]
We now pass to the reciprocals. From the relation
\[ x^4=o(2^x) \]
it follows that \(2^x\) grows more rapidly than \(x^4\). Consequently, the reciprocal of \(2^x\) tends to zero more rapidly than the reciprocal of \(x^4\):
\[ \frac{1}{2^x} = o\left(\frac{1}{x^4}\right). \]
Indeed,
\[ \frac{\displaystyle\frac{1}{2^x}} {\displaystyle\frac{1}{x^4}} = \frac{x^4}{2^x} \to 0. \]
Similarly, from the relation
\[ \ln(x)=o(\sqrt{x}) \]
it follows that
\[ \frac{1}{\sqrt{x}} = o\left(\frac{1}{\ln(x)}\right). \]
Indeed,
\[ \frac{\displaystyle\frac{1}{\sqrt{x}}} {\displaystyle\frac{1}{\ln(x)}} = \frac{\ln(x)}{\sqrt{x}} \to 0. \]
Exercise 20 — level ★★★★★
Evaluate the limit
\[ \lim_{x\to 0} \frac{(1-\cos x)\ln(1+x)} {x^2\sin x}. \]
Answer
\[ \lim_{x\to 0} \frac{(1-\cos x)\ln(1+x)} {x^2\sin x} = \frac{1}{2}. \]
Solution
The limit presents the indeterminate form \(\displaystyle\frac{0}{0}\), since as \(x\to 0\),
\[ 1-\cos x\to 0, \qquad \ln(1+x)\to 0, \qquad x^2\sin x\to 0. \]
The numerator is a product, and so is the denominator. We may therefore use the asymptotic equivalences of the individual factors.
As \(x\to 0\),
\[ 1-\cos x\sim\frac{x^2}{2}. \]
Moreover,
\[ \ln(1+x)\sim x \]
and
\[ \sin x\sim x. \]
We replace each factor with its equivalent:
\[ \lim_{x\to 0} \frac{(1-\cos x)\ln(1+x)} {x^2\sin x} = \lim_{x\to 0} \frac{\displaystyle\frac{x^2}{2}\cdot x} {x^2\cdot x}. \]
We multiply the factors in the numerator:
\[ \frac{x^2}{2}\cdot x=\frac{x^3}{2}. \]
We multiply the factors in the denominator:
\[ x^2\cdot x=x^3. \]
The limit therefore becomes
\[ \lim_{x\to 0}\frac{\displaystyle\frac{x^3}{2}}{x^3}. \]
Dividing \(\displaystyle\frac{x^3}{2}\) by \(x^3\) is equivalent to writing
\[ \frac{x^3}{2x^3}. \]
For \(x\neq 0\), we may cancel \(x^3\):
\[ \frac{x^3}{2x^3}=\frac{1}{2}. \]
Hence
\[ \lim_{x\to 0} \frac{(1-\cos x)\ln(1+x)} {x^2\sin x} = \frac{1}{2}. \]