The following practice problems are designed to develop, step by step, an understanding of the following consequence of the definition of limit: if a function has a non-zero limit, then, sufficiently near the limiting point, it has the same sign as its limit. We begin with direct applications, in which it suffices to evaluate a limit and determine its sign, and proceed to consequences such as the non-vanishing of a function, the signs of products and quotients, and comparisons between functions.
Exercise 1 — level ★☆☆☆☆
Let
\[ f(x)=x+3. \]
Using the permanence-of-sign theorem, determine the sign of \(f(x)\) on a suitable neighbourhood of \(x=0\).
Answer
There is a neighbourhood of \(0\) on which
\[ f(x)>0. \]
Solution
To apply the permanence-of-sign theorem, we must first compute the limit of the function at the point in question.
Since \(f(x)=x+3\), we have
\[ \lim_{x\to0}(x+3)=3. \]
The value of the limit is therefore
\[ L=3>0. \]
This result states that, whenever the limit is strictly positive, the function itself is positive for all \(x\) sufficiently close to the limiting point.
Hence there exists \(\delta>0\) such that
\[ f(x)>0 \]
for every \(x\in(-\delta,\delta)\setminus\{0\}\).
In this simple case we can also identify a suitable neighbourhood directly. Indeed,
\[ x+3>0 \]
is equivalent to
\[ x>-3. \]
For instance, taking \(\delta=1\), if \(-1<x<1\) then certainly \(x>-3\), so that
\[ x+3>0. \]
We have thus verified concretely the conclusion guaranteed by the theorem.
Exercise 2 — level ★☆☆☆☆
Let
\[ f(x)=2x-5. \]
Determine the sign of \(f(x)\) on a suitable neighbourhood of \(x=1\).
Answer
There is a neighbourhood of \(1\) on which
\[ f(x)<0. \]
Solution
We first compute the limit of the function as \(x\to1\):
\[ \lim_{x\to1}(2x-5)=2\cdot1-5=-3. \]
The limit is therefore
\[ L=-3<0. \]
Since the limit is strictly negative, the permanence-of-sign theorem guarantees that \(f(x)\) is negative for all \(x\) sufficiently close to \(1\).
There thus exists \(\delta>0\) such that
\[ 2x-5<0 \]
for every \(x\in(1-\delta,1+\delta)\setminus\{1\}\).
We can verify this conclusion directly. The inequality
\[ 2x-5<0 \]
is equivalent to
\[ x<\frac52. \]
It therefore suffices to choose a neighbourhood of \(1\) that lies entirely to the left of \(\displaystyle\frac{5}{2}\). For instance, with \(\delta=1\) we get
\[ 0<x<2, \]
and hence \(x<\displaystyle\frac{5}{2}\). Thus \(f(x)<0\) on that neighbourhood.
Exercise 3 — level ★☆☆☆☆
Consider
\[ f(x)=x^2+1. \]
Use the permanence-of-sign theorem to determine the sign of the function on a neighbourhood of \(x=2\).
Answer
There is a neighbourhood of \(2\) on which
\[ f(x)>0. \]
Solution
We compute the limit:
\[ \lim_{x\to2}(x^2+1)=2^2+1=5. \]
We therefore have
\[ L=5>0. \]
Since the limit is strictly positive, the permanence-of-sign theorem ensures that there is a neighbourhood of \(2\) on which \(f(x)\) is likewise positive.
Hence, for a suitable \(\delta>0\),
\[ x^2+1>0 \]
for every \(x\in(2-\delta,2+\delta)\setminus\{2\}\).
In this case we can in fact observe something stronger: since \(x^2\geq0\) for every \(x\in\mathbb R\), we have
\[ x^2+1\geq1>0 \]
for every real number \(x\). The function is thus positive not merely near \(2\), but on the whole of \(\mathbb R\).
The theorem, however, only guarantees a local conclusion: from the positivity of the limit we may certainly deduce the positivity of the function on a suitable neighbourhood of \(2\).
Exercise 4 — level ★☆☆☆☆
Let
\[ f(x)=\frac{x+1}{x+2}. \]
Determine the sign of \(f(x)\) on a suitable neighbourhood of \(x=0\).
Answer
There is a neighbourhood of \(0\) on which
\[ f(x)>0. \]
Solution
We compute the limit as \(x\to0\). Since the denominator tends to \(2\neq0\), we may substitute \(x=0\) directly:
\[ \lim_{x\to0}\frac{x+1}{x+2} = \frac{1}{2}. \]
The value of the limit is
\[ L=\frac12>0. \]
The permanence-of-sign theorem then allows us to conclude that the function takes positive values for all \(x\) sufficiently close to \(0\).
Symbolically, there exists \(\delta>0\) such that
\[ \frac{x+1}{x+2}>0 \]
for every \(x\in(-\delta,\delta)\setminus\{0\}\).
It is worth noting that we had no need to solve a rational inequality directly: the positive sign of the limit alone suffices to guarantee the positive sign of the function on a suitable neighbourhood of the point.
Exercise 5 — level ★☆☆☆☆
Consider
\[ f(x)=\ln(x)-1. \]
Determine the sign of \(f(x)\) on a suitable neighbourhood of \(x=1\).
Answer
There is a neighbourhood of \(1\) on which
\[ f(x)<0. \]
Solution
The function is defined for \(x>0\), and hence in particular on a sufficiently small neighbourhood of \(1\).
We compute the limit:
\[ \lim_{x\to1}(\ln(x)-1)=\ln(1)-1=0-1=-1. \]
We thus have
\[ L=-1<0. \]
By this result, there exists \(\delta>0\) such that, if
\[ 0<|x-1|<\delta, \]
then
\[ \ln(x)-1<0. \]
In other words, the function has the same sign as its limit and is negative on a suitable neighbourhood of \(1\).
Exercise 6 — level ★★☆☆☆
Given that
\[ \lim_{x\to x_0}f(x)=4, \]
prove that there is a neighbourhood of \(x_0\) on which
\[ f(x)>2. \]
Answer
There exists \(\delta>0\) such that
\[ f(x)>2 \]
for every \(x\in(x_0-\delta,x_0+\delta)\setminus\{x_0\}\).
Solution
We wish to prove not merely that \(f(x)\) is positive near \(x_0\), but the sharper property
\[ f(x)>2. \]
We know that
\[ \lim_{x\to x_0}f(x)=4. \]
By the definition of limit, for every \(\epsilon>0\) there exists \(\delta>0\) such that
\[ |f(x)-4|<\epsilon \]
whenever \(0<|x-x_0|<\delta\).
We want the resulting lower bound to be precisely \(2\). We therefore choose
\[ \epsilon=2. \]
The definition of limit then guarantees the existence of \(\delta>0\) such that
\[ |f(x)-4|<2. \]
Removing the absolute value gives
\[ -2<f(x)-4<2. \]
Adding \(4\) throughout:
\[ 2<f(x)<6. \]
In particular,
\[ f(x)>2. \]
We have thus obtained not merely the positivity of \(f(x)\), but a sharper quantitative estimate.
Exercise 7 — level ★★☆☆☆
Given that
\[ \lim_{x\to x_0}f(x)=-6, \]
prove that there is a neighbourhood of \(x_0\) on which
\[ f(x)<-3. \]
Answer
For \(x\) sufficiently close to \(x_0\),
\[ f(x)<-3. \]
Solution
We know that \(f(x)\) tends to \(-6\). We wish to show that, for \(x\) sufficiently close to \(x_0\), the values of \(f(x)\) remain below \(-3\).
By the definition of limit, for any chosen \(\epsilon>0\), for \(x\) sufficiently close to \(x_0\) we have
\[ |f(x)+6|<\epsilon. \]
We choose
\[ \epsilon=3. \]
We obtain
\[ |f(x)+6|<3. \]
This is equivalent to
\[ -3<f(x)+6<3. \]
Subtracting \(6\) throughout:
\[ -9<f(x)<-3. \]
From this double inequality it follows immediately that
\[ f(x)<-3. \]
In particular, \(f(x)\) is negative on a suitable neighbourhood of \(x_0\), exactly as predicted by the permanence-of-sign theorem.
Exercise 8 — level ★★☆☆☆
Let
\[ \lim_{x\to x_0}f(x)=L \]
with \(L>0\). Prove that there is a neighbourhood of \(x_0\) on which
\[ f(x)>\frac{L}{2}. \]
Answer
There exists \(\delta>0\) such that, if \(0<|x-x_0|<\delta\), then
\[ f(x)>\frac{L}{2}>0. \]
Solution
Since \(L>0\), we also have
\[ \frac{L}{2}>0. \]
In the definition of limit we choose
\[ \epsilon=\frac{L}{2}. \]
There thus exists \(\delta>0\) such that, if
\[ 0<|x-x_0|<\delta, \]
then
\[ |f(x)-L|<\frac{L}{2}. \]
Removing the absolute value:
\[ -\frac{L}{2}<f(x)-L<\frac{L}{2}. \]
Adding \(L\) throughout:
\[ \frac{L}{2}<f(x)<\frac{3L}{2}. \]
In particular,
\[ f(x)>\frac{L}{2}. \]
Since \(\displaystyle\frac{L}{2}>0\), it also follows that
\[ f(x)>0. \]
This is precisely the positive case in the proof of the permanence-of-sign theorem.
Exercise 9 — level ★★☆☆☆
Let
\[ \lim_{x\to x_0}f(x)=L \]
with \(L<0\). Prove that, for \(x\) sufficiently close to \(x_0\),
\[ f(x)<\frac{L}{2}<0. \]
Answer
There exists \(\delta>0\) such that
\[ f(x)<\frac{L}{2}<0 \]
for every \(x\) with \(0<|x-x_0|<\delta\).
Solution
Since \(L<0\), we have
\[ |L|=-L. \]
We choose
\[ \epsilon=\frac{|L|}{2}=-\frac{L}{2}. \]
This quantity is positive, so it is an admissible choice in the definition of limit.
For \(x\) sufficiently close to \(x_0\) we then have
\[ |f(x)-L|<\frac{|L|}{2}. \]
From this inequality we obtain in particular
\[ f(x)<L+\frac{|L|}{2}. \]
Since \(|L|=-L\),
\[ L+\frac{|L|}{2} = L-\frac{L}{2} = \frac{L}{2}. \]
Hence
\[ f(x)<\frac{L}{2}. \]
Since \(L<0\), we also have \(\displaystyle\frac{L}{2}<0\). Consequently
\[ f(x)<\frac{L}{2}<0, \]
and the function is negative on a suitable neighbourhood of \(x_0\).
Exercise 10 — level ★★☆☆☆
Prove that, if
\[ \lim_{x\to x_0}f(x)=L\neq0, \]
then \(f(x)\neq0\) for all \(x\) sufficiently close to \(x_0\).
Answer
There is a neighbourhood of \(x_0\) on which
\[ f(x)\neq0. \]
Solution
The key hypothesis is
\[ L\neq0. \]
A non-zero real number is either positive or negative. We must therefore distinguish two cases.
If
\[ L>0, \]
the permanence-of-sign theorem guarantees that, for \(x\) sufficiently close to \(x_0\),
\[ f(x)>0. \]
A strictly positive number cannot be zero, so
\[ f(x)\neq0. \]
If instead
\[ L<0, \]
the same theorem guarantees
\[ f(x)<0 \]
for \(x\) sufficiently close to \(x_0\). Here too \(f(x)\) cannot equal zero.
In either case there is thus a neighbourhood of \(x_0\) on which
\[ f(x)\neq0. \]
This consequence is particularly important whenever \(f(x)\) appears in the denominator of a quotient.
Exercise 11 — level ★★★☆☆
Let \(f\) and \(g\) be two functions such that
\[ \lim_{x\to x_0}f(x)=2, \qquad \lim_{x\to x_0}g(x)=-3. \]
Determine the sign of \(f(x)g(x)\) for \(x\) sufficiently close to \(x_0\).
Answer
For \(x\) sufficiently close to \(x_0\),
\[ f(x)g(x)<0. \]
Solution
We examine the two functions separately.
Since
\[ \lim_{x\to x_0}f(x)=2>0, \]
by the permanence-of-sign theorem there is a neighbourhood of \(x_0\) on which
\[ f(x)>0. \]
Likewise,
\[ \lim_{x\to x_0}g(x)=-3<0, \]
so there is a neighbourhood of \(x_0\) on which
\[ g(x)<0. \]
We may choose a sufficiently small neighbourhood on which both properties hold simultaneously.
On that neighbourhood \(f(x)\) is positive and \(g(x)\) is negative. The product of a positive and a negative number is negative. Hence
\[ f(x)g(x)<0. \]
Exercise 12 — level ★★★☆☆
Let \(f\) and \(g\) be such that
\[ \lim_{x\to x_0}f(x)=-4, \qquad \lim_{x\to x_0}g(x)=-2. \]
Determine the sign of the product \(f(x)g(x)\) near \(x_0\).
Answer
For \(x\) sufficiently close to \(x_0\),
\[ f(x)g(x)>0. \]
Solution
The limit of \(f(x)\) is
\[ -4<0. \]
By the permanence of sign, \(f(x)<0\) for \(x\) sufficiently close to \(x_0\).
The limit of \(g(x)\) is likewise negative:
\[ -2<0. \]
Hence, again by the permanence-of-sign theorem,
\[ g(x)<0 \]
on a suitable neighbourhood of \(x_0\).
By shrinking the neighbourhood if necessary, we may arrange for both inequalities to hold simultaneously.
The product of two negative numbers is positive. It therefore follows that
\[ f(x)g(x)>0 \]
for \(x\) sufficiently close to \(x_0\).
Exercise 13 — level ★★★☆☆
Let \(f\) and \(g\) be two functions such that
\[ \lim_{x\to x_0}f(x)=5, \qquad \lim_{x\to x_0}g(x)=-2. \]
Determine the sign of the quotient
\[ \frac{f(x)}{g(x)} \]
for \(x\) sufficiently close to \(x_0\).
Answer
For \(x\) sufficiently close to \(x_0\),
\[ \frac{f(x)}{g(x)}<0. \]
Solution
From the first hypothesis we have
\[ \lim_{x\to x_0}f(x)=5>0. \]
The permanence-of-sign theorem then implies that
\[ f(x)>0 \]
for \(x\) sufficiently close to \(x_0\).
For the second function,
\[ \lim_{x\to x_0}g(x)=-2<0. \]
It follows that
\[ g(x)<0 \]
for \(x\) sufficiently close to \(x_0\).
In particular \(g(x)\neq0\) on that neighbourhood, so the quotient is well defined.
Since the numerator is positive and the denominator is negative, their quotient is negative:
\[ \frac{f(x)}{g(x)}<0. \]
Exercise 14 — level ★★★☆☆
Let
\[ \lim_{x\to x_0}f(x)=-3. \]
Determine the sign of the reciprocal function
\[ \frac{1}{f(x)} \]
for \(x\) sufficiently close to \(x_0\).
Answer
For \(x\) sufficiently close to \(x_0\),
\[ \frac{1}{f(x)}<0. \]
Solution
We have
\[ \lim_{x\to x_0}f(x)=-3<0. \]
By the permanence-of-sign theorem there is thus a neighbourhood of \(x_0\) on which
\[ f(x)<0. \]
Since \(f(x)\) is strictly negative there, it cannot equal zero on that neighbourhood. Consequently the reciprocal function
\[ \frac{1}{f(x)} \]
is well defined.
The reciprocal of a negative number is again negative. Hence
\[ \frac{1}{f(x)}<0 \]
for \(x\) sufficiently close to \(x_0\).
Exercise 15 — level ★★★☆☆
Consider
\[ f(x)=x^2. \]
As \(x\to0\) we have
\[ \lim_{x\to0}x^2=0. \]
Can the permanence-of-sign theorem be applied directly here? What can nonetheless be said about the sign of \(f(x)\) on a neighbourhood of \(0\)?
Answer
The permanence-of-sign theorem, in the form requiring a limit \(L\neq0\), does not apply. Nevertheless
\[ x^2>0 \]
for every \(x\neq0\).
Solution
The limit of the function is
\[ \lim_{x\to0}x^2=0. \]
The version of the permanence-of-sign theorem we have studied requires the limit to be non-zero:
\[ L\neq0. \]
In this exercise, however,
\[ L=0. \]
We cannot therefore use the theorem directly to deduce the sign of the function.
This does not mean that \(f(x)\) fails to have a definite sign near \(0\); it simply means that we must establish it by other means.
For every real number \(x\),
\[ x^2\geq0. \]
Moreover,
\[ x^2=0 \]
only when \(x=0\). Hence, on a neighbourhood of \(0\), for \(x\neq0\) we always have
\[ x^2>0. \]
This exercise illustrates an important point: having a non-zero limit is a sufficient condition for applying the permanence-of-sign theorem, but a function may nonetheless retain a fixed sign even when its limit equals zero.
Exercise 16 — level ★★★☆☆
Consider
\[ f(x)=x \]
as \(x\to0\). Determine whether the function retains a constant sign on a neighbourhood of \(0\).
Answer
No. In every neighbourhood of \(0\), the function takes both negative and positive values.
Solution
The limit is
\[ \lim_{x\to0}x=0. \]
Here too the limit is zero, so the permanence-of-sign theorem, in the form studied, does not apply.
Let us examine the behaviour of the function directly.
If \(x<0\), then
\[ f(x)=x<0. \]
If instead \(x>0\), then
\[ f(x)=x>0. \]
Every neighbourhood of \(0\) contains both negative and positive numbers. Consequently, in every neighbourhood of \(0\), the function takes values of both signs.
There is thus no neighbourhood of \(0\) on which \(f(x)\) is always positive or always negative.
Comparing this exercise with the previous one, we see that when the limit equals zero it is not possible to determine the sign of the function from the value of the limit alone: \(x^2\) is positive on a neighbourhood of \(0\), whereas \(x\) changes sign.
Exercise 17 — level ★★★★☆
Let
\[ f_a(x)=x^2+a, \]
where \(a\in\mathbb R\). Using the permanence-of-sign theorem wherever possible, study the sign of \(f_a(x)\) on a neighbourhood of \(x=0\) in the cases \(a>0\), \(a<0\) and \(a=0\).
Answer
If \(a>0\), then \(f_a(x)>0\) near \(0\).
If \(a<0\), then \(f_a(x)<0\) near \(0\).
If \(a=0\), the theorem does not apply directly, but \(f_0(x)=x^2>0\) for every \(x\neq0\).
Solution
We first compute the limit as a function of the parameter \(a\):
\[ \lim_{x\to0}(x^2+a)=a. \]
The sign of the limit therefore depends directly on the value of \(a\).
If
\[ a>0, \]
the limit is strictly positive. By the permanence-of-sign theorem there is thus a neighbourhood of \(0\) on which
\[ f_a(x)=x^2+a>0. \]
If instead
\[ a<0, \]
the limit is strictly negative. The theorem then guarantees that, for \(x\) sufficiently close to \(0\),
\[ f_a(x)=x^2+a<0. \]
This last conclusion may seem less immediate, since \(x^2\) is always non-negative. However, when \(x\) is very close to \(0\), \(x^2\) is also very small and is not enough to offset the negative constant term \(a\).
There remains the case
\[ a=0. \]
In this case
\[ \lim_{x\to0}f_0(x)=0. \]
This result does not apply directly, because the limit is zero. Nevertheless,
\[ f_0(x)=x^2>0 \]
for every \(x\neq0\).
This exercise thus makes clear the role played by the hypothesis \(L\neq0\).
Exercise 18 — level ★★★★☆
Let \(f\) and \(g\) be two functions such that
\[ \lim_{x\to x_0}f(x)=7, \qquad \lim_{x\to x_0}g(x)=3. \]
Prove that, for \(x\) sufficiently close to \(x_0\),
\[ f(x)>g(x). \]
Answer
There is a neighbourhood of \(x_0\) on which
\[ f(x)>g(x). \]
Solution
To compare \(f(x)\) and \(g(x)\), it is convenient to consider their difference:
\[ h(x)=f(x)-g(x). \]
Saying that
\[ f(x)>g(x) \]
is indeed equivalent to saying that
\[ f(x)-g(x)>0. \]
We therefore compute the limit of the difference:
\[ \lim_{x\to x_0}[f(x)-g(x)] = \lim_{x\to x_0}f(x)-\lim_{x\to x_0}g(x). \]
Substituting the given values:
\[ \lim_{x\to x_0}[f(x)-g(x)] = 7-3 = 4. \]
Since
\[ 4>0, \]
the permanence-of-sign theorem, applied to the function \(h(x)=f(x)-g(x)\), guarantees that
\[ f(x)-g(x)>0 \]
for \(x\) sufficiently close to \(x_0\).
Adding \(g(x)\) to both sides we obtain
\[ f(x)>g(x). \]
Exercise 19 — level ★★★★☆
Let \(f\) and \(g\) be two functions such that
\[ \lim_{x\to x_0}f(x)=-2, \qquad \lim_{x\to x_0}g(x)=3. \]
Determine, for \(x\) sufficiently close to \(x_0\), the sign of each of the expressions
\[ f(x),\qquad g(x),\qquad f(x)g(x),\qquad \frac{f(x)}{g(x)}. \]
Answer
For \(x\) sufficiently close to \(x_0\),
\[ f(x)<0,\qquad g(x)>0,\qquad f(x)g(x)<0,\qquad \frac{f(x)}{g(x)}<0. \]
Solution
We begin with the function \(f\). Since
\[ \lim_{x\to x_0}f(x)=-2<0, \]
the permanence-of-sign theorem guarantees that
\[ f(x)<0 \]
for \(x\) sufficiently close to \(x_0\).
For the function \(g\), on the other hand,
\[ \lim_{x\to x_0}g(x)=3>0. \]
Hence
\[ g(x)>0 \]
on a suitable neighbourhood of \(x_0\).
By shrinking the neighbourhood if necessary, we may assume that both properties hold simultaneously.
Since \(f(x)\) is negative and \(g(x)\) is positive, their product is negative:
\[ f(x)g(x)<0. \]
Moreover \(g(x)>0\), so in particular
\[ g(x)\neq0. \]
The quotient is therefore well defined. Since the numerator is negative and the denominator is positive, we obtain
\[ \frac{f(x)}{g(x)}<0. \]
We have thus used this result to determine simultaneously the signs of several expressions built from the two functions.
Exercise 20 — level ★★★★★
Let \(f\), \(g\) and \(h\) be three functions such that
\[ \lim_{x\to x_0}f(x)=-2, \qquad \lim_{x\to x_0}g(x)=5, \qquad \lim_{x\to x_0}h(x)=-1. \]
Determine the sign of the expression
\[ \frac{[f(x)-g(x)]h(x)}{g(x)} \]
for \(x\) sufficiently close to \(x_0\).
Answer
For \(x\) sufficiently close to \(x_0\),
\[ \frac{[f(x)-g(x)]h(x)}{g(x)}>0. \]
Solution
The expression contains several components, so we proceed systematically and determine the sign of each part in turn.
We begin with the difference
\[ f(x)-g(x). \]
Using the algebra of limits:
\[ \lim_{x\to x_0}[f(x)-g(x)] = -2-5 = -7. \]
Since
\[ -7<0, \]
the permanence-of-sign theorem guarantees that
\[ f(x)-g(x)<0 \]
for \(x\) sufficiently close to \(x_0\).
Consider now \(h(x)\). We have
\[ \lim_{x\to x_0}h(x)=-1<0. \]
Consequently,
\[ h(x)<0 \]
near \(x_0\).
The numerator of the expression is thus the product
\[ [f(x)-g(x)]h(x) \]
of two negative quantities. The product of two negative numbers is positive, so
\[ [f(x)-g(x)]h(x)>0. \]
We turn finally to the denominator. Since
\[ \lim_{x\to x_0}g(x)=5>0, \]
the permanence-of-sign theorem implies
\[ g(x)>0 \]
for \(x\) sufficiently close to \(x_0\).
In particular \(g(x)\neq0\), so the quotient is well defined.
We thus have a positive numerator and a positive denominator. Their quotient is positive:
\[ \frac{[f(x)-g(x)]h(x)}{g(x)}>0. \]
It is worth noting that each individual sign property might initially hold only on some particular neighbourhood of \(x_0\). Since there are only finitely many such neighbourhoods to consider, however, we may always choose a single, sufficiently small neighbourhood on which all the properties obtained hold simultaneously.