Evaluating Limits with Powers, Logarithms, and Exponentials

Welcome back. In the previous lesson, you evaluated a proper definite integral by finding an antiderivative and applying

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

Improper integrals use the same antiderivative method, but one of the limits may be $b\to \infty$, $a\to -\infty$, or $x\to a^{+}$. Therefore, your ability to evaluate limits accurately is essential.

This lesson builds a compact limit toolkit for powers, logarithms, and exponentials. By the end, you should be able to recognize the relevant form, calculate the limit using the quickest method, and decide whether the result is finite or infinite.

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1. What does “as $x$ approaches infinity” mean?

The notation

$$\underset{x\to \infty }{\lim }f(x)$$

does not mean that we substitute a number called infinity into $f(x)$. Infinity is not an ordinary number. It means: what happens to $f(x)$ when $x$ becomes arbitrarily large and positive?

Similarly,

$$\underset{x\to -\infty }{\lim }f(x)$$

asks what happens when $x$ becomes arbitrarily large in the negative direction.

For improper integrals, a limit at infinity often appears after evaluating an antiderivative. For example, later you will write an integral over an infinite interval using a temporary upper limit $b$:

$${\int }_1^{\infty }f(x)dx=\underset{b\to \infty }{\lim }{\int }_1^{b}f(x)dx.$$

The integral has a finite value only if this final limit exists and is finite.

Three possible outcomes

A limit may:

1. Approach a finite number, such as $0$, $2$, or $-5$.
2. Increase without bound:

$$\underset{x\to \infty }{\lim }f(x)=\infty .$$

3. Decrease without bound:

$$\underset{x\to \infty }{\lim }f(x)=-\infty .$$

An infinite result is not a finite limit. In improper-integral questions, it normally signals divergence.

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2. Limits involving powers

For $x>0$, the basic power limits are worth memorising.

As $x\to \infty$

For any $p>0$,

$$\underset{x\to \infty }{\lim }{x}^p=\infty ,$$

while

$$\underset{x\to \infty }{\lim }\frac{1}{x^p}=\underset{x\to \infty }{\lim }{x}^{-p}=0.$$

A positive power grows without bound. A negative power is a reciprocal, so it tends to zero.

For example,

$$\underset{x\to \infty }{\lim }\frac{7}{x^3}=0.$$

As $x\to 0^{+}$

The behavior reverses:

$$\underset{x\to 0^{+}}{\lim }{x}^p=0$$

and

$$\underset{x\to 0^{+}}{\lim }\frac{1}{x^p}=\infty .$$

For instance,

$$\underset{x\to 0^{+}}{\lim }\frac{1}{\sqrt{x}}=\underset{x\to 0^{+}}{\lim }{x}^{-1/2}=\infty .$$

This second limit is especially important. It shows that $\frac{1}{\sqrt{x}}$ becomes unbounded close to $x=0$, even though it can still sometimes have a convergent improper integral. We will examine that distinction later.

Powers at negative infinity

For integer powers, the parity of the exponent matters:

$$\underset{x\to -\infty }{\lim }{x}^{\text{even}}=\infty ,$$

while

$$\underset{x\to -\infty }{\lim }{x}^{\text{odd}}=-\infty .$$

Thus,

$$\underset{x\to -\infty }{\lim }{x}^4=\infty ,\underset{x\to -\infty }{\lim }{x}^3=-\infty .$$

A negative coefficient reverses the sign. Hence,

$$\underset{x\to -\infty }{\lim }(-2x^4)=-\infty .$$

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3. Dominant terms: polynomials and rational functions

When $x$ is very large, the highest power of $x$ dominates a polynomial.

For example,

$$5x^4-3x^2+7$$

behaves like $5x^4$ as $x\to \infty$, because $x^4$ becomes much larger than $x^2$ and the constant $7$.

Therefore,

$$\underset{x\to \infty }{\lim }(5x^4-3x^2+7)=\infty .$$

For rational functions, compare the degrees of the numerator and denominator.

Degree comparison	Limit as $x\to \infty$
Numerator degree $<$ denominator degree	$0$
Numerator degree $=$ denominator degree	Ratio of leading coefficients
Numerator degree $>$ denominator degree	Usually $\infty$ or $-\infty$; inspect leading terms

Worked example 1: Denominator has higher degree

$$\underset{x\to \infty }{\lim }\frac{4x^2-1}{3x^5+2x}$$

The denominator has degree $5$, while the numerator has degree $2$. Therefore,

$$\boxed{\underset{x\to \infty }{\lim }\frac{4x^2-1}{3x^5+2x}=0.}$$

For full exam working, divide every term by $x^5$:

$$\frac{4x^2-1}{3x^5+2x}=\frac{\frac{4}{x^3}-\frac{1}{x^5}}{3+\frac{2}{x^4}}.$$

As $x\to \infty$, every reciprocal power tends to $0$:

$$\underset{x\to \infty }{\lim }\frac{\frac{4}{x^3}-\frac{1}{x^5}}{3+\frac{2}{x^4}}=\frac{0-0}{3+0}=0.$$

Worked example 2: Equal degrees

$$\underset{x\to \infty }{\lim }\frac{6x^3-x+4}{2x^3+5x^2}.$$

The degrees are both $3$. Take the ratio of leading coefficients:

$$\boxed{\underset{x\to \infty }{\lim }\frac{6x^3-x+4}{2x^3+5x^2}=\frac{6}{2}=3.}$$

Worked example 3: Numerator has higher degree

$$\underset{x\to \infty }{\lim }\frac{2x^4-x}{5x^2+1}.$$

The leading-term behavior is

$$\frac{2x^4}{5x^2}=\frac{2}{5}{x}^2.$$

Since $x^2\to \infty$,

$$\boxed{\underset{x\to \infty }{\lim }\frac{2x^4-x}{5x^2+1}=\infty .}$$

For limits at infinity, “keep the leading terms” is a fast method. Dividing by the highest denominator power is the safer full-working method.

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4. Natural logarithms near $0$ and infinity

The natural logarithm $\ln x$ is defined only when

$$x>0.$$

Its two essential limits are:

$$\underset{x\to 0^{+}}{\lim }\ln x=-\infty$$

and

$$\underset{x\to \infty }{\lim }\ln x=\infty .$$

The notation $x\to 0^{+}$ means that $x$ approaches $0$ through positive values only. This is necessary because $\ln x$ is not defined for negative $x$.

Although $\ln x$ tends to infinity as $x\to \infty$, it grows very slowly. Any positive power of $x$ grows faster.

For $p>0$,

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x^p}=0.$$

For example,

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x^2}=0.$$

This is an $\frac{\infty }{\infty }$ indeterminate form, so L’Hopital’s rule is allowed:

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x^2}=\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{2x}.$$

Simplify:

$$\frac{\frac{1}{x}}{2x}=\frac{1}{2x^2}.$$

Thus,

$$\boxed{\underset{x\to \infty }{\lim }\frac{\ln x}{x^2}=0.}$$

A related result near zero is also extremely useful:

$$\underset{x\to 0^{+}}{\lim }{x}^p\ln x=0(p>0).$$

Even though $\ln x\to -\infty$, the positive power $x^p\to 0$ strongly enough that the product approaches $0$.

For example,

$$\underset{x\to 0^{+}}{\lim }{x}^2\ln x$$

has the indeterminate form $0\cdot (-\infty )$. Rewrite it as a quotient:

$$x^2\ln x=\frac{\ln x}{x^{-2}}.$$

Now it has the form $\frac{-\infty }{\infty }$, so apply L’Hopital’s rule:

$$\underset{x\to 0^{+}}{\lim }\frac{\ln x}{x^{-2}}=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{-2x^{-3}}.$$ $$=\underset{x\to 0^{+}}{\lim }\left(-\frac{x^2}{2}\right)=0.$$

Therefore,

$$\boxed{\underset{x\to 0^{+}}{\lim }{x}^2\ln x=0.}$$

Do not conclude that every product involving $0$ and infinity equals $0$. The expression $0\cdot \infty$ is indeterminate. It must be rewritten and evaluated.

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5. Exponential limits

The natural exponential function has the following end behavior:

$$\underset{x\to \infty }{\lim }{e}^x=\infty ,\underset{x\to -\infty }{\lim }{e}^x=0.$$

For the decaying exponential $e^{-x}$, the behavior is reversed:

$$\underset{x\to \infty }{\lim }{e}^{-x}=0,\underset{x\to -\infty }{\lim }{e}^{-x}=\infty .$$

A useful way to remember this is:

$$e^{-x}=\frac{1}{e^x}.$$

As $x\to \infty$, the denominator $e^x$ becomes extremely large, so the fraction tends to $0$.

The key growth hierarchy for $x\to \infty$ is

$$\ln x\text{grows more slowly than}{x}^p\text{grows more slowly than}{e}^{ax},$$

where $p>0$ and $a>0$.

In limit form:

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x^p}=0,$$ $$\underset{x\to \infty }{\lim }\frac{x^p}{e^{ax}}=0.$$

Worked example 4: Exponential dominates a power

Evaluate

$$\underset{x\to \infty }{\lim }\frac{x^3}{e^x}.$$

Both numerator and denominator tend to infinity, giving the indeterminate form $\frac{\infty }{\infty }$. Apply L’Hopital’s rule repeatedly:

$$\underset{x\to \infty }{\lim }\frac{x^3}{e^x}=\underset{x\to \infty }{\lim }\frac{3x^2}{e^x}.$$ $$=\underset{x\to \infty }{\lim }\frac{6x}{e^x}.$$ $$=\underset{x\to \infty }{\lim }\frac{6}{e^x}.$$

Since $e^x\to \infty$,

$$\boxed{\underset{x\to \infty }{\lim }\frac{x^3}{e^x}=0.}$$

A standard finite limit involving $e^x$

At $x=0$,

$$\underset{x\to 0}{\lim }\frac{e^x-1}{x}$$

has the form $\frac{0}{0}$, because $e^0-1=0$. L’Hopital’s rule gives

$$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=\underset{x\to 0}{\lim }\frac{e^x}{1}.$$ $$=e^0=1.$$

Hence,

$$\boxed{\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1.}$$

This limit is particularly useful when exponential expressions occur near a finite endpoint.

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6. When to use L’Hopital’s rule

L’Hopital’s rule is powerful, but it is not the first method for every limit.

You may apply it directly only when substitution produces one of these indeterminate quotient forms:

$$\frac{0}{0}\text{or}\frac{\infty }{\infty }.$$

Then differentiate the numerator and denominator separately:

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)},$$

provided the conditions for the rule are satisfied and the resulting limit exists.

Do not use L’Hopital’s rule immediately for:

$$\frac{5}{\infty },\frac{\infty }{4},0\cdot \infty ,\infty -\infty .$$

The first two are already understandable:

$$\frac{5}{\infty }\text{ behaves like }0,\frac{\infty }{4}\text{ behaves like }\infty .$$

For $0\cdot \infty$, rewrite the product as a quotient. For $\infty -\infty$, combine terms or simplify algebraically first.

Efficient exam decision rule

When you see a limit, work in this order:

1. Substitute mentally and identify the form.
2. For polynomials and rational functions at infinity, use dominant powers.
3. For logarithms and exponentials, use the growth hierarchy:

$$\ln x\ll x^p\ll e^{ax}.$$

4. Use L’Hopital’s rule only if the expression is $\frac{0}{0}$ or $\frac{\infty }{\infty }$, or after valid rewriting into one of those forms.
5. State the final result clearly as a finite number, $0$, $\infty$, or $-\infty$.

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7. Connection to improper integrals

Here is how these limits will appear in the next lesson.

Suppose an antiderivative produces $e^{-x}$:

$$\underset{b\to \infty }{\lim }{\left[-e^{-x}\right]}_1^b.$$

First apply upper minus lower:

$$=\underset{b\to \infty }{\lim }\left(-e^{-b}+e^{-1}\right).$$

Since

$$\underset{b\to \infty }{\lim }{e}^{-b}=0,$$

the result is finite:

$$\underset{b\to \infty }{\lim }\left(-e^{-b}+e^{-1}\right)=e^{-1}.$$

In contrast, if an antiderivative produces $\ln b$, then

$$\underset{b\to \infty }{\lim }\ln b=\infty .$$

That result is not finite, so the corresponding improper integral will diverge.

At this stage, focus on the limit itself. The next lesson will show the complete definition and notation for integrals over infinite intervals.

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Key takeaways

Keep these facts in your notes:

$$\underset{x\to \infty }{\lim }{x}^p=\infty (p>0),$$ $$\underset{x\to \infty }{\lim }\frac{1}{x^p}=0(p>0),$$ $$\underset{x\to 0^{+}}{\lim }\frac{1}{x^p}=\infty (p>0),$$ $$\underset{x\to 0^{+}}{\lim }\ln x=-\infty ,\underset{x\to \infty }{\lim }\ln x=\infty ,$$ $$\underset{x\to \infty }{\lim }{e}^{-x}=0,\underset{x\to -\infty }{\lim }{e}^x=0,$$ $$\underset{x\to \infty }{\lim }\frac{\ln x}{x^p}=0,\underset{x\to \infty }{\lim }\frac{x^p}{e^{ax}}=0(p,a>0),$$ $$\underset{x\to 0^{+}}{\lim }{x}^p\ln x=0(p>0).$$

Most importantly: do not treat $\infty$ as a number to substitute. Identify the form, simplify using dominant behavior where possible, and use L’Hopital’s rule only for valid indeterminate quotients.

Next, you will use these limits to rewrite and evaluate Type 1 improper integrals, where the interval of integration extends to $\infty$ or $-\infty$.