\[ \lim \limits _{t \to \infty } \frac {x^6}{e^x} = \answer {0} \]
relatively bigger
\(\rhd \) Do functions tend to infinity differently?
For example,
Both \(p(x) = 6x^3 - 5x + 2\) and \(w(t) = 2t^2 = 4t - 5\) approach \(\infty \) as you move further out in the domain.
\[ \lim _{x \to \infty } p(x) = \infty \, \text { and } \, \lim _{t \to \infty } w(t) = \infty \]
\(\blacktriangleright \) Do they approach \(\infty \) in the same way?
\(\blacktriangleright \) What do we mean by “same way”?
How should we compare \(p(x)\) and \(w(t)\) to decide if they approach \(\infty \) in the same way?
We have two ways of comparing things: differences and quotients.
Quotients work better for comparing the end-behavior of functions.
If two functions approach \(\infty \) in the “same way”, then their quotient should approach a constant as you move out in the domain.
By the “same way”, we are picturing a horizontal asymptote for the quotient.
For instance, rational functions approach \(\infty \) when the degrees of the polynomials are equal. The polynomial in the numerator and the polynomial in the denominator “behave in the same way” as we move far out in the domain. The two polynomials are somehow “at the same level” of infinity.
To decide if two functions, \(f\) and \(g\), approach infinity in the same way,
- (a)
- create a quotient: \(\frac {f(x)}{g(x)}\)
- (b)
- examine the end-behavior of this quotient: \(\lim \limits _{x \to \infty } \frac {f(x)}{g(x)}\)
- (c)
- if this limit equals a nonzero constant, then \(f\) and \(g\) approach \(\infty \) the same - neither overshadows the other.
If the limit is \(1\), then people sometimes say the two functions are asymptotically equivalent. People sometimes use the symbol \(\sim \) for asymptotically equivalent.
Perhaps more important for future Calculus courses is when this limit is \(0\).
Dominance
Dominance
If the quotient limit equals \(0\),
\[ \lim _{x \to \infty } \frac {f(x)}{g(x)} = 0\]
then we say that \(g\) dominates \(f\).
Sometimes people use a double inequality for dominates: \(f \ll g\).
When one function dominates another, then it approaches infinity in such a way that the other function is like \(0\) in comparison. When the dominate function is the denominator, then this drives the quotient to \(0\).
Our initial order of dominance looks like this.
Order of Dominance
When we are comparing unbounded function values, then
\[ bounded functions (constants, \, sine, \, cosine) \ll logarithmic \ll power (polynomial) \ll exponential \]
This goes for their powers as well.
\[ bounded functions (constants, \, sine, \, cosine) \ll logarithmic^{power} \ll power polynomial^{power} \ll exponential \]
Note: power functions include roots and radicals and linear and polynomials.
Dominance
![[Picture]](dominance-4384dc13c2ffde92d90073731bd9ae26.svg)
![[Picture]](dominance-2e341145ced61c63fc6b623c4c1d557d.svg)
Let \(g(x) = \frac {\ln (x)}{x}\).
Graph of \(y = g(x)\).
According to our order of dominance, the denominator is a linear function and should dominate the numerator function, which is a logarithm. Therefore, as \(x\) approach infinity, this whole fraction should approach \(0\).
Let’s zoom out.
Here we can see the graph peak and begin to come back down.
However, we are investigating end-behavior. \(18\) is not very large. We are interested in the values of \(g(x)\) when \(x\) is very very large.
That was not a good graph for our purposes. We are interested in large values of \(x\).
Eventually, the graph approaches the x-axis as the function value approaches \(0\).
The denominator function, \(x\), dominates the numerator function, \(\ln (x)\). The values of \(\ln (x)\) become like \(0\) compared to the values of \(x\), when \(x\) is very very very large.
Dominance is not when \(x\) takes on small values.
Dominance is not when \(x\) has medium values.
Dominance is a statement when \(x\) takes on very very very large values.
The level of dominance includes powers. We can raise the power of \(\ln (x)\), but they are still overshadowed by values of \(x\), when \(x\) is very very very large.
Powers
Graph of \(y = \frac {(\ln (x))^5}{x}\).
\(x\) should dominate any power of \(\ln (x)\). Therefore, the graph of this function should eventually aproach the the \(x\)-axis as the function value approaches \(0\).
Thinking: If this function approaches \(0\), then it must get below and stay below \(2\), eventually. How far out in the domain do you need to go before this function eventually dips and stays below \(2\).
Hint: Look out around \(1,000,000\).
For the following limits, refer to our order of dominance.
Upsidedown
When we talk about dominance, we phrase it in terms of the quotient approaching \(0\). The quotient might be approaching \(0\) through positive numbers or negative numbers. It doesn’t matter because \(-0 = 0\).
If we flip our quotient upsidedown and the dominate function is in the numerator, then the quotient becomes unbounded. Then we also need to think about the sign.
\[ \lim \limits _{k \to \infty } \frac {8}{3^k} = 0 \]
\[ \lim \limits _{k \to \infty } \frac {3^k}{8} = \infty \]
Since \(4 - k \ll 3^k\),
\[ \lim \limits _{k \to \infty } \frac {4 - k}{3^k} = 0 \]
When thinking about the reciprocal, then we need to keep in mind that \(4 - k < 0\) for large positive values of \(k\).
\[ \lim \limits _{k \to \infty } \frac {3^k}{4 - k} = -\infty \]
Sums and Differences
Dominance also helps us identify the most important terms in a function’s formula.
When one term in a formula dominates over an added term, then we can remove the dominated term from our analysis.
Consider the function \(B(t) = \frac {3e^t - 5}{e^t + 1}\).
\(\blacktriangleright \) As \(t\) becomes unbounded positively, \(3e^t\) dominates \(-5\).
\(\blacktriangleright \) As \(t\) becomes unbounded positively, \(3^t\) dominates \(1\).
As \(t\) becomes unbounded positively,
\[ B(t) = \frac {3e^t - 5}{e^t + 1} = \frac {2 e^t}{e^t} \cdot \frac {3 - \tfrac {5}{e^t}}{1 + \tfrac {1}{e^t}} \sim \frac {2 e^t}{e^t} \]
As \(t\) becomes unbounded positively, the exponential pieces will dominate the constant pieces in each expression.
\[ B(t) = \frac {3e^t - 5}{e^t + 1} \sim \frac {3 e^t}{e^t} = 3 \]
\(\blacktriangleright \) Now consider the other direction.
As \(t\) becomes unbounded negatively, we have a different story.
\[ \lim \limits _{t \to -\infty } 3 e^t = 0 \]
This reverses the dominance.
\[ \lim \limits _{t \to -\infty } \frac {e^t}{5} = 0 \]
As \(t\) becomes unbounded negatively, we have a different story. \(\lim \limits _{t \to -\infty } e^t = 0\). Therefore, in this case, the constant terms dominate.
\[ B(t) = \frac {3e^t - 5}{e^t + 1} \sim \frac {-5}{1} = -5 \]
This helps us think about limiting behavior.
\[ \lim \limits _{x \to -\infty } \frac {e^{x} + 5x^2 + \sqrt {4-x} - 7}{5 + 6x^4 + e^{-x}} = \lim \limits _{x \to -\infty } \frac {5x^2}{e^{-x}} = 0 \]
\[ \lim \limits _{x \to \infty } \frac {\sqrt {x^3 - x + 1}}{x \, \sqrt {2 \, x-6}} = \lim \limits _{x \to \infty } \frac {\sqrt {x^3}}{x \, \sqrt {2 \, x}} = \lim \limits _{x \to \infty } \frac {x \, \sqrt {x}}{x \, \sqrt {x} \, \sqrt {2} } = \lim \limits _{x \to \infty } \frac {1}{\sqrt {2} } = \frac {1}{\sqrt {2} } \]
ooooo-=-=-=-ooOoo-=-=-=-ooooo
more examples can be found by following this link
More Examples of Dominance