In this collection we present 20 practice problems on the hierarchy of infinite growth, arranged in order of increasing difficulty. The problems cover the comparison of growth rates, functions of higher and lower order, functions of the same order, asymptotic equivalence, the fundamental growth hierarchies, and the identification of the dominant term.
Each problem is solved step by step, highlighting the criterion used and explaining why it applies.
Problem 1 — level ★☆☆☆☆
Compare the growth of \(x^2\) and \(x^5\) as \(x\to+\infty\) and determine which of the two is of higher order.
Answer
We have
\[ x^2=o\bigl(x^5\bigr) \qquad\text{as }x\to+\infty. \]
Hence \(x^2\) is of lower order than \(x^5\).
Solution
Both functions tend to \(+\infty\) as \(x\to+\infty\). To compare their rates of growth, consider the ratio of the first to the second:
\[ \frac{x^2}{x^5}. \]
Using the properties of powers, this simplifies to:
\[ \frac{x^2}{x^5}=\frac{1}{x^3}. \]
As \(x\to+\infty\), \(x^3\to+\infty\) as well, so its reciprocal tends to \(0\):
\[ \frac{1}{x^3}\to0. \]
We have therefore obtained
\[ \lim_{x\to+\infty}\frac{x^2}{x^5}=0. \]
By definition, this means that \(x^2\) is negligible compared with \(x^5\). In little-o notation,
\[ x^2=o\bigl(x^5\bigr). \]
Consequently, \(x^2\) is of lower order than \(x^5\), while \(x^5\) is of higher order than \(x^2\).
Problem 2 — level ★☆☆☆☆
Determine whether \(4x^3\) and \(7x^3\) are of the same order as \(x\to+\infty\), and whether they are equivalent.
Answer
The functions \(4x^3\) and \(7x^3\) are of the same order, but they are not equivalent.
Solution
To compare the two functions, we compute the limit of their ratio:
\[ \lim_{x\to+\infty}\frac{4x^3}{7x^3}. \]
Since \(x^3\neq0\) for \(x\) sufficiently large, we may cancel the factor \(x^3\):
\[ \frac{4x^3}{7x^3}=\frac{4}{7}. \]
Hence
\[ \lim_{x\to+\infty}\frac{4x^3}{7x^3}=\frac{4}{7}. \]
The limit is a positive finite real number. This is precisely the criterion that characterises two functions of the same order.
For the two functions to be equivalent as well, the limit of the ratio would need to equal \(1\). Here, instead,
\[ \frac{4}{7}\neq1. \]
Therefore \(4x^3\) and \(7x^3\) are of the same order, but they are not equivalent.
Problem 3 — level ★☆☆☆☆
Show that
\[ x^4+2x^2\sim x^4 \qquad\text{as }x\to+\infty. \]
Answer
We have
\[ x^4+2x^2\sim x^4. \]
Solution
To verify an asymptotic equivalence we must examine the ratio of the two functions and check whether it tends to \(1\).
So consider
\[ \frac{x^4+2x^2}{x^4}. \]
Divide each term of the numerator separately by \(x^4\):
\[ \frac{x^4+2x^2}{x^4} = 1+\frac{2}{x^2}. \]
As \(x\to+\infty\),
\[ \frac{2}{x^2}\to0. \]
Consequently,
\[ \lim_{x\to+\infty} \frac{x^4+2x^2}{x^4} = 1. \]
By the definition of asymptotic equivalence, we may therefore conclude that
\[ x^4+2x^2\sim x^4. \]
This result also expresses the fact that, in the sum \(x^4+2x^2\), the term \(x^4\) is dominant: the term \(2x^2\) grows more slowly and is negligible compared with \(x^4\).
Problem 4 — level ★★☆☆☆
Compare \(x^3\) and \(x^4\) as \(x\to-\infty\).
Answer
We have
\[ x^3=o\bigl(x^4\bigr) \qquad\text{as }x\to-\infty. \]
Hence \(x^3\) is of lower order than \(x^4\).
Solution
Here it is important to note that, as \(x\to-\infty\),
\[ x^3\to-\infty, \qquad x^4\to+\infty. \]
The two functions therefore have opposite signs. To compare their magnitudes correctly, consider the ratio of their absolute values:
\[ \frac{|x^3|}{|x^4|}. \]
Using the properties of absolute value,
\[ |x^3|=|x|^3, \qquad |x^4|=|x|^4. \]
Hence
\[ \frac{|x^3|}{|x^4|} = \frac{|x|^3}{|x|^4} = \frac{1}{|x|}. \]
As \(x\to-\infty\), we have \(|x|\to+\infty\). Consequently,
\[ \frac{1}{|x|}\to0. \]
It follows that
\[ \lim_{x\to-\infty} \frac{|x^3|}{|x^4|} =0. \]
The magnitude of \(x^3\) is therefore negligible compared with that of \(x^4\). Equivalently,
\[ x^3=o\bigl(x^4\bigr) \qquad\text{as }x\to-\infty. \]
Problem 5 — level ★★☆☆☆
Compare \(\ln(x)\) and \(x^{1/3}\) as \(x\to+\infty\).
Answer
We have
\[ \ln(x)=o\bigl(x^{1/3}\bigr). \]
Solution
Both functions tend to \(+\infty\), but they belong to two different families: \(\ln(x)\) is a logarithmic function, whereas \(x^{1/3}\) is a positive power of \(x\).
The fundamental hierarchy of growth rates states that any positive power of the natural logarithm grows more slowly than any positive power of \(x\):
\[ \bigl(\ln(x)\bigr)^p=o\bigl(x^q\bigr) \]
for every \(p>0\) and \(q>0\).
In our case we may take
\[ p=1, \qquad q=\frac{1}{3}. \]
We obtain immediately
\[ \ln(x)=o\bigl(x^{1/3}\bigr). \]
In terms of the ratio, this relation means
\[ \lim_{x\to+\infty} \frac{\ln(x)}{x^{1/3}} =0. \]
Thus, even though both functions become arbitrarily large, \(x^{1/3}\) grows asymptotically faster than \(\ln(x)\).
Problem 6 — level ★★☆☆☆
Evaluate
\[ \lim_{x\to+\infty} \frac{\bigl(\ln(x)\bigr)^4}{\sqrt{x}}. \]
Answer
\[ 0 \]
Solution
First, write the square root as a power:
\[ \sqrt{x}=x^{1/2}. \]
The limit then becomes
\[ \lim_{x\to+\infty} \frac{\bigl(\ln(x)\bigr)^4}{x^{1/2}}. \]
The fundamental hierarchy states that, for every \(p>0\) and \(q>0\),
\[ \bigl(\ln(x)\bigr)^p=o\bigl(x^q\bigr). \]
Taking
\[ p=4, \qquad q=\frac{1}{2}, \]
we obtain
\[ \bigl(\ln(x)\bigr)^4=o\bigl(x^{1/2}\bigr). \]
By the definition of little-o, this means precisely that
\[ \lim_{x\to+\infty} \frac{\bigl(\ln(x)\bigr)^4}{x^{1/2}} =0. \]
Therefore
\[ \lim_{x\to+\infty} \frac{\bigl(\ln(x)\bigr)^4}{\sqrt{x}} =0 . \]
Problem 7 — level ★★☆☆☆
Evaluate
\[ \lim_{x\to+\infty}\frac{x^6}{e^x}. \]
Answer
\[ 0 \]
Solution
The numerator \(x^6\) is a positive power of \(x\), while the denominator \(e^x\) is an exponential function with base \(e>1\).
The fundamental hierarchy of growth rates states that every positive power of \(x\) grows more slowly than every exponential with base greater than \(1\):
\[ x^q=o\bigl(a^x\bigr) \]
for every \(q>0\) and \(a>1\).
In our case
\[ q=6, \qquad a=e. \]
Hence
\[ x^6=o\bigl(e^x\bigr). \]
The little-o notation means precisely that the ratio of \(x^6\) to \(e^x\) tends to zero:
\[ \lim_{x\to+\infty}\frac{x^6}{e^x}=0. \]
Problem 8 — level ★★☆☆☆
Compare \(3^x\) and \(7^x\) as \(x\to+\infty\).
Answer
\[ 3^x=o\bigl(7^x\bigr). \]
Solution
Both functions are exponentials with base greater than \(1\). Within this family, the exponential with the larger base grows faster.
We can verify this directly by examining the ratio:
\[ \frac{3^x}{7^x}. \]
Using the property of powers with the same exponent,
\[ \frac{3^x}{7^x} = \left(\frac{3}{7}\right)^x. \]
Since
\[ 0<\frac{3}{7}<1, \]
the power \(\left(\displaystyle\frac{3}{7}\right)^x\) tends to \(0\) as \(x\to+\infty\). Hence
\[ \lim_{x\to+\infty} \frac{3^x}{7^x} =0. \]
By definition,
\[ 3^x=o\bigl(7^x\bigr). \]
Hence \(3^x\) is of lower order than \(7^x\).
Problem 9 — level ★★☆☆☆
Compare
\[ \ln\bigl(\ln(x)\bigr) \qquad\text{and}\qquad \sqrt{\ln(x)} \]
as \(x\to+\infty\).
Answer
\[ \ln\bigl(\ln(x)\bigr) = o\bigl(\sqrt{\ln(x)}\bigr). \]
Solution
Write the second function as a power of the logarithm:
\[ \sqrt{\ln(x)} = \bigl(\ln(x)\bigr)^{1/2}. \]
The hierarchy of growth rates extends to the iterated logarithm and gives
\[ \ln\bigl(\ln(x)\bigr) = o\bigl((\ln(x))^p\bigr) \]
for every \(p>0\).
Taking
\[ p=\frac{1}{2}, \]
we obtain
\[ \ln\bigl(\ln(x)\bigr) = o\bigl((\ln(x))^{1/2}\bigr). \]
Since \((\ln(x))^{1/2}=\sqrt{\ln(x)}\), we may conclude that
\[ \ln\bigl(\ln(x)\bigr) = o\bigl(\sqrt{\ln(x)}\bigr). \]
Problem 10 — level ★★☆☆☆
Compare \(e^x\) and \(x^x\) as \(x\to+\infty\).
Answer
\[ e^x=o\bigl(x^x\bigr). \]
Solution
Both functions tend to \(+\infty\), but \(x^x\) grows faster than any exponential \(a^x\) with fixed base \(a>1\).
Indeed, the extended fundamental hierarchy contains the relation
\[ a^x=o\bigl(x^x\bigr) \]
for every fixed base \(a>1\).
Taking \(a=e\), we obtain immediately
\[ e^x=o\bigl(x^x\bigr). \]
Equivalently,
\[ \lim_{x\to+\infty}\frac{e^x}{x^x}=0. \]
Hence \(x^x\) is of higher order than \(e^x\).
Problem 11 — level ★★★☆☆
Arrange the following functions, from the slowest-growing to the fastest-growing, as \(x\to+\infty\):
\[ x^4,\qquad 2^x,\qquad \bigl(\ln(x)\bigr)^3,\qquad x^{1/2},\qquad x^x,\qquad \ln\bigl(\ln(x)\bigr). \]
Answer
The correct order is
\[ \ln\bigl(\ln(x)\bigr), \quad \bigl(\ln(x)\bigr)^3, \quad x^{1/2}, \quad x^4, \quad 2^x, \quad x^x. \]
Solution
We proceed by identifying the family to which each function belongs.
The function
\[ \ln\bigl(\ln(x)\bigr) \]
is an iterated logarithm. It grows more slowly than any positive power of \(\ln(x)\). In particular,
\[ \ln\bigl(\ln(x)\bigr) = o\bigl((\ln(x))^3\bigr). \]
Next we compare \((\ln(x))^3\) with \(x^{1/2}\). Every positive power of the logarithm grows more slowly than every positive power of \(x\), so
\[ (\ln(x))^3=o\bigl(x^{1/2}\bigr). \]
Between the two powers \(x^{1/2}\) and \(x^4\), the one with the larger exponent grows faster:
\[ x^{1/2}=o\bigl(x^4\bigr). \]
Every positive power of \(x\) grows more slowly than an exponential with base greater than \(1\), so
\[ x^4=o\bigl(2^x\bigr). \]
Finally, every exponential with fixed base greater than \(1\) grows more slowly than \(x^x\):
\[ 2^x=o\bigl(x^x\bigr). \]
Putting all these comparisons together, we obtain the chain
\[ \ln\bigl(\ln(x)\bigr) = o\bigl((\ln(x))^3\bigr), \]
\[ (\ln(x))^3 = o\bigl(x^{1/2}\bigr), \]
\[ x^{1/2} = o\bigl(x^4\bigr), \]
\[ x^4=o\bigl(2^x\bigr), \qquad 2^x=o\bigl(x^x\bigr). \]
Hence, from the slowest-growing to the fastest-growing, the required order is
\[ \ln\bigl(\ln(x)\bigr), \qquad \bigl(\ln(x)\bigr)^3, \qquad x^{1/2}, \qquad x^4, \qquad 2^x, \qquad x^x. \]
Problem 12 — level ★★★☆☆
Determine the dominant term of
\[ f(x)=5x^6-2x^3+7 \]
as \(x\to+\infty\), and find a function equivalent to \(f\).
Answer
The dominant term is \(5x^6\), and
\[ 5x^6-2x^3+7\sim5x^6. \]
Solution
Among positive powers of \(x\), the one with the larger exponent grows faster. In the function
\[ 5x^6-2x^3+7 \]
the term with the highest power is \(5x^6\).
To prove rigorously that it is dominant, we divide the whole function by \(5x^6\):
\[ \frac{5x^6-2x^3+7}{5x^6} = 1-\frac{2x^3}{5x^6}+\frac{7}{5x^6}. \]
Simplifying,
\[ \frac{5x^6-2x^3+7}{5x^6} = 1-\frac{2}{5x^3}+\frac{7}{5x^6}. \]
As \(x\to+\infty\),
\[ \frac{2}{5x^3}\to0, \qquad \frac{7}{5x^6}\to0. \]
Hence
\[ \lim_{x\to+\infty} \frac{5x^6-2x^3+7}{5x^6} =1. \]
By the definition of asymptotic equivalence,
\[ 5x^6-2x^3+7\sim5x^6 . \]
Problem 13 — level ★★★☆☆
Determine the dominant term of
\[ f(x)=4e^x+x^8-\bigl(\ln(x)\bigr)^5 \]
as \(x\to+\infty\).
Answer
The dominant term is \(4e^x\), and
\[ 4e^x+x^8-\bigl(\ln(x)\bigr)^5\sim4e^x. \]
Solution
The three terms belong to different families:
\[ \bigl(\ln(x)\bigr)^5, \qquad x^8, \qquad e^x. \]
The fundamental hierarchy gives
\[ \bigl(\ln(x)\bigr)^5=o\bigl(x^8\bigr) \]
and
\[ x^8=o\bigl(e^x\bigr). \]
The exponential is therefore the fastest-growing term.
To verify that the full expression is indeed equivalent to \(4e^x\), we divide by \(4e^x\):
\[ \frac{4e^x+x^8-(\ln(x))^5}{4e^x} = 1+\frac{x^8}{4e^x} - \frac{(\ln(x))^5}{4e^x}. \]
Since \(x^8=o(e^x)\),
\[ \frac{x^8}{4e^x}\to0. \]
For the term \((\ln(x))^5\) we use both comparisons from the hierarchy. Indeed, we can write
\[ \frac{(\ln(x))^5}{4e^x} = \frac{1}{4} \cdot \frac{(\ln(x))^5}{x^8} \cdot \frac{x^8}{e^x}. \]
Since
\[ \frac{(\ln(x))^5}{x^8}\to0 \qquad\text{and}\qquad \frac{x^8}{e^x}\to0, \]
their product also tends to \(0\). Hence
\[ \frac{(\ln(x))^5}{4e^x}\to0. \]
We thus obtain
\[ \lim_{x\to+\infty} \frac{4e^x+x^8-(\ln(x))^5}{4e^x} =1. \]
Hence
\[ 4e^x+x^8-\bigl(\ln(x)\bigr)^5\sim4e^x . \]
Problem 14 — level ★★★☆☆
Evaluate
\[ \lim_{x\to+\infty} \frac{(x^3+2x)(\ln(x)+1)} {x^3\ln(x)}. \]
Answer
\[ 1 \]
Solution
The numerator is the product of two factors. We study each of them separately.
For the first factor,
\[ \frac{x^3+2x}{x^3} = 1+\frac{2}{x^2}\to1. \]
Hence
\[ x^3+2x\sim x^3. \]
For the second factor,
\[ \frac{\ln(x)+1}{\ln(x)} = 1+\frac{1}{\ln(x)}. \]
Since \(\ln(x)\to+\infty\),
\[ \frac{1}{\ln(x)}\to0. \]
Hence
\[ \ln(x)+1\sim\ln(x). \]
Asymptotic equivalence is preserved under products. Hence
\[ (x^3+2x)(\ln(x)+1) \sim x^3\ln(x). \]
By the definition of equivalence,
\[ \lim_{x\to+\infty} \frac{(x^3+2x)(\ln(x)+1)} {x^3\ln(x)} =1. \]
Problem 15 — level ★★★☆☆
Determine whether
\[ 3x^4+x \qquad\text{and}\qquad 5x^4-2 \]
are of the same order, or equivalent, as \(x\to+\infty\).
Answer
The two functions are of the same order, but they are not equivalent.
Solution
Consider the ratio
\[ \frac{3x^4+x}{5x^4-2}. \]
Since the highest-degree term in both the numerator and the denominator is of degree four, we divide both by \(x^4\):
\[ \frac{3x^4+x}{5x^4-2} = \frac{3+\displaystyle\frac{1}{x^3}}{5-\displaystyle\frac{2}{x^4}}. \]
As \(x\to+\infty\),
\[ \frac{1}{x^3}\to0, \qquad \frac{2}{x^4}\to0. \]
Hence
\[ \lim_{x\to+\infty} \frac{3x^4+x}{5x^4-2} = \frac{3}{5}. \]
The limit is positive and finite, so the two functions are of the same order.
However,
\[ \frac{3}{5}\neq1. \]
Consequently they are not equivalent.
We may also observe separately that
\[ 3x^4+x\sim3x^4, \qquad 5x^4-2\sim5x^4, \]
and that the two dominant terms differ by a constant factor.
Problem 16 — level ★★★★☆
Evaluate
\[ \lim_{x\to+\infty} \frac{e^x+x^6}{e^x-3x^2}. \]
Answer
\[ 1 \]
Solution
The numerator contains \(e^x\) and \(x^6\). The denominator contains \(e^x\) and \(x^2\).
The fundamental hierarchy tells us that any positive power of \(x\) is negligible compared with \(e^x\). In particular,
\[ x^6=o\bigl(e^x\bigr), \qquad x^2=o\bigl(e^x\bigr). \]
The dominant term in both the numerator and the denominator is therefore \(e^x\).
Divide the numerator and the denominator by \(e^x\):
\[ \frac{e^x+x^6}{e^x-3x^2} = \frac{1+\displaystyle\frac{x^6}{e^x}}{1-\displaystyle\frac{3x^2}{e^x}}. \]
By the relations above,
\[ \frac{x^6}{e^x}\to0 \]
and
\[ \frac{3x^2}{e^x}\to0. \]
Passing to the limit, we obtain
\[ \lim_{x\to+\infty} \frac{1+\displaystyle\frac{x^6}{e^x}}{1-\displaystyle\frac{3x^2}{e^x}} = \frac{1+0}{1-0} =1. \]
Hence
\[ \lim_{x\to+\infty} \frac{e^x+x^6}{e^x-3x^2} =1 . \]
We may express the same result by writing
\[ e^x+x^6\sim e^x \qquad\text{and}\qquad e^x-3x^2\sim e^x. \]
Problem 17 — level ★★★★☆
Given that
\[ \ln(x)=o\bigl(\sqrt{x}\bigr) \qquad\text{as }x\to+\infty, \]
determine the corresponding relation between the reciprocals
\[ \frac{1}{\ln(x)} \qquad\text{and}\qquad \frac{1}{\sqrt{x}}. \]
Answer
We have
\[ \frac{1}{\sqrt{x}} = o\left(\frac{1}{\ln(x)}\right). \]
Solution
The initial relation
\[ \ln(x)=o\bigl(\sqrt{x}\bigr) \]
means, by definition, that
\[ \lim_{x\to+\infty} \frac{\ln(x)}{\sqrt{x}} =0. \]
Now we wish to compare the reciprocals
\[ \frac{1}{\sqrt{x}} \qquad\text{and}\qquad \frac{1}{\ln(x)}. \]
Form the ratio of the first to the second:
\[ \frac{\displaystyle\frac{1}{\sqrt{x}}}{\displaystyle\frac{1}{\ln(x)}}. \]
Dividing by a fraction is the same as multiplying by its reciprocal, so
\[ \frac{\displaystyle\frac{1}{\sqrt{x}}}{\displaystyle\frac{1}{\ln(x)}} = \frac{\ln(x)}{\sqrt{x}}. \]
But we have already established that
\[ \frac{\ln(x)}{\sqrt{x}}\to0. \]
Consequently,
\[ \lim_{x\to+\infty} \frac{\displaystyle\frac{1}{\sqrt{x}}}{\displaystyle\frac{1}{\ln(x)}} =0. \]
By the definition of little-o,
\[ \frac{1}{\sqrt{x}} = o\left(\frac{1}{\ln(x)}\right) . \]
Passing to reciprocals therefore reverses the order of magnitude: the function that grows faster has the reciprocal that tends to zero faster.
Problem 18 — level ★★★★☆
Given that
\[ x^2+x\sim x^2 \qquad\text{as }x\to+\infty, \]
explain why it is not correct to replace \(x^2+x\) with \(x^2\) in the difference
\[ (x^2+x)-x^2, \]
and evaluate
\[ \lim_{x\to+\infty} \frac{(x^2+x)-x^2}{x}. \]
Answer
In this case it is not legitimate to replace \(x^2+x\) with its equivalent \(x^2\) in the difference, because cancelling the dominant terms would cause the term \(x\) to be lost.
The limit equals
\[ 1. \]
Solution
We begin by verifying the stated equivalence:
\[ \frac{x^2+x}{x^2} = 1+\frac{1}{x}\to1. \]
So it is indeed true that
\[ x^2+x\sim x^2. \]
However, asymptotic equivalence is not an equality. Saying that \(x^2+x\sim x^2\) does not mean that we may always replace \(x^2+x\) with \(x^2\) in every expression.
Indeed, consider the exact difference:
\[ (x^2+x)-x^2. \]
The terms \(x^2\) cancel, leaving
\[ (x^2+x)-x^2=x. \]
Consequently,
\[ \frac{(x^2+x)-x^2}{x} = \frac{x}{x} =1. \]
Hence
\[ \lim_{x\to+\infty} \frac{(x^2+x)-x^2}{x} =1 . \]
Let us now see what would happen if we improperly carried out the substitution
\[ x^2+x\sim x^2 \]
directly in the difference. We would obtain
\[ x^2-x^2=0, \]
losing entirely the term \(x\), which is in fact exactly what remains after the cancellation.
This example shows why equivalents may be used directly in products and quotients, but not indiscriminately in sums and differences.
Problem 19 — level ★★★★☆
Evaluate
\[ \lim_{x\to+\infty} \frac{(\ln(x))^3+x^4}{e^x}. \]
Answer
\[ 0 \]
Solution
The numerator is the sum of two functions:
\[ (\ln(x))^3 \qquad\text{and}\qquad x^4. \]
Let us first compare them with each other. By the fundamental hierarchy,
\[ (\ln(x))^3=o\bigl(x^4\bigr). \]
Hence, within the numerator, the dominant term is \(x^4\).
We can verify this by dividing the numerator by \(x^4\):
\[ \frac{(\ln(x))^3+x^4}{x^4} = \frac{(\ln(x))^3}{x^4}+1. \]
Since
\[ \frac{(\ln(x))^3}{x^4}\to0, \]
it follows that
\[ (\ln(x))^3+x^4\sim x^4. \]
Now compare \(x^4\) with the denominator \(e^x\). A positive power of \(x\) grows more slowly than an exponential:
\[ x^4=o\bigl(e^x\bigr). \]
We may also split the original ratio directly:
\[ \frac{(\ln(x))^3+x^4}{e^x} = \frac{(\ln(x))^3}{e^x} + \frac{x^4}{e^x}. \]
The second term tends to \(0\) because \(x^4=o(e^x)\).
For the first term we may write
\[ \frac{(\ln(x))^3}{e^x} = \frac{(\ln(x))^3}{x^4} \cdot \frac{x^4}{e^x}. \]
Both factors tend to \(0\), so their product also tends to \(0\).
Consequently,
\[ \lim_{x\to+\infty} \frac{(\ln(x))^3+x^4}{e^x} =0 . \]
Problem 20 — level ★★★★★
Evaluate
\[ \lim_{x\to+\infty} \frac{x^x+3^x}{2x^x+x^5}. \]
Answer
\[ \frac{1}{2} \]
Solution
This limit involves three different types of growth:
\[ x^5,\qquad 3^x,\qquad x^x. \]
We must first determine which function dominates the others.
The fundamental hierarchy gives us
\[ x^5=o\bigl(3^x\bigr), \]
because every positive power of \(x\) grows more slowly than every exponential with base greater than \(1\).
Furthermore,
\[ 3^x=o\bigl(x^x\bigr), \]
because every exponential with fixed base greater than \(1\) grows more slowly than \(x^x\).
The dominant term in both the numerator and the denominator is therefore \(x^x\).
Divide the numerator and the denominator by \(x^x\):
\[ \frac{x^x+3^x}{2x^x+x^5} = \frac{1+\displaystyle\frac{3^x}{x^x}}{2+\displaystyle\frac{x^5}{x^x}}. \]
From the relation
\[ 3^x=o\bigl(x^x\bigr) \]
it follows immediately that
\[ \frac{3^x}{x^x}\to0. \]
We must now also justify carefully that
\[ \frac{x^5}{x^x}\to0. \]
We use the two comparisons from the hierarchy already established:
\[ \frac{x^5}{3^x}\to0 \qquad\text{and}\qquad \frac{3^x}{x^x}\to0. \]
We may write
\[ \frac{x^5}{x^x} = \frac{x^5}{3^x} \cdot \frac{3^x}{x^x}. \]
Both factors tend to \(0\). Hence their product tends to \(0\):
\[ \frac{x^5}{x^x}\to0. \]
We may now pass to the limit in the expression obtained:
\[ \lim_{x\to+\infty} \frac{1+\displaystyle\frac{3^x}{x^x}}{2+\displaystyle\frac{x^5}{x^x}} = \frac{1+0}{2+0} = \frac{1}{2}. \]
Hence
\[ \lim_{x\to+\infty} \frac{x^x+3^x}{2x^x+x^5} = \frac{1}{2} . \]
The result may also be interpreted through the dominant terms:
\[ x^x+3^x\sim x^x \]
and
\[ 2x^x+x^5\sim2x^x. \]
The ratio between the two functions therefore behaves like
\[ \frac{x^x}{2x^x}=\frac{1}{2}. \]