Logarithmic Limits of Indeterminate Forms

What Makes an Exponential Form Indeterminate?

Before learning the technique, it's crucial to build an intuition for why these forms are uncertain. In arithmetic, $1^n=1$ and $x^0=1$. But in calculus, we are dealing with limits, not fixed numbers. The "1", "0", and "$\infty$" represent functions that are approaching these values. The final result depends on the rate of their approach.

Consider the form $1^{\infty }$. It describes a limit of the type $\lim f(x{)}^{g(x)}$ where $f(x)\to 1$ and $g(x)\to \infty$.

• The base, $f(x)$, is getting closer and closer to 1, which tends to make the whole expression 1.
• The exponent, $g(x)$, is growing infinitely large, which tends to pull the expression away from 1. If the base is even slightly greater than 1 (e.g., 1.00001), an infinite exponent would make it explode to $\infty$. If it's slightly less than 1 (e.g., 0.99999), it would shrink to 0.

Who wins this tug-of-war? We can't know without more information. The expression is indeterminate. The same logic applies to $0^0$ and ${\infty }^0$.

The following video from Mathologer gives a fantastic conceptual overview of what it means for a form to be indeterminate. Pay close attention to how he frames the problem in terms of limits of functions.

The General Method: Logarithmic Transformation

So, how do we resolve this uncertainty? We can't evaluate $\lim f(x{)}^{g(x)}$ directly. The trick is to bring the exponent $g(x)$ down from its perch so we can handle it. The properties of logarithms are perfect for this.

The general strategy is as follows:

1. Let the limit you want to find be $L$. So, $L={\lim }_{x\to a}f(x{)}^{g(x)}$.
2. To make things simpler, first set $y=f(x{)}^{g(x)}$.
3. Take the natural logarithm of both sides: $\ln (y)=\ln (f(x{)}^{g(x)})$.
4. Using the logarithm power rule, we get: $\ln (y)=g(x)\cdot \ln (f(x))$.
5. Now, find the limit of this new expression: ${\lim }_{x\to a}\ln (y)={\lim }_{x\to a}[g(x)\cdot \ln (f(x))]$. Let's say this limit evaluates to $K$.
6. We have found that $\ln (L)=K$. To find $L$, we simply exponentiate: $L=e^K$.

This process transforms an exponential indeterminate form ($1^{\infty },0^0,{\infty }^0$) into a product indeterminate form ($\infty \cdot 0$), which is a step in the right direction.

The following text explains this general procedure. While it uses a tool called L'Hôpital's rule that we will learn later, the initial setup of the logarithmic transformation is universal and perfectly described.

Tackling $1^{\infty }$: A Powerful Shortcut

The $1^{\infty }$ form is the most common one you'll encounter on the JEE. For this specific case, the general logarithmic method can be simplified into a very fast and elegant formula.

Let's start with our transformed expression's limit: $K={\lim }_{x\to a}[g(x)\cdot \ln (f(x))]$.
Since ${\lim }_{x\to a}f(x)=1$, we can write $f(x)=1+h(x)$, where ${\lim }_{x\to a}h(x)=0$.
Substituting this in, we get:
$K={\lim }_{x\to a}[g(x)\cdot \ln (1+h(x))]$
Now for a clever trick. Let's multiply and divide by $h(x)$:
$K={\lim }_{x\to a}\left[g(x)\cdot h(x)\cdot \frac{\ln (1+h(x))}{h(x)}\right]$
Look at the fraction on the right. As $x\to a$, $h(x)\to 0$. This is exactly one of our standard limit forms from the previous lesson!
${\lim }_{h(x)\to 0}\frac{\ln (1+h(x))}{h(x)}=1$
So, our expression for $K$ simplifies beautifully:
$K={\lim }_{x\to a}[g(x)\cdot h(x)\cdot 1]={\lim }_{x\to a}[g(x)\cdot (f(x)-1)]$
Since the original limit was $e^K$, we arrive at the final formula:

If a limit is of the form $1^{\infty }$, ${\lim }_{x\to a}f(x{)}^{g(x)}$ can be calculated as:

This formula converts the $1^{\infty }$ problem directly into evaluating a simpler limit in the exponent. This is a must-know shortcut for competitive exams.

The following resource provides an excellent derivation of this formula and several worked examples.

Let's work through one more example together: ${\lim }_{x\to \infty }{\left(\frac{x+3}{x-1}\right)}^{x+2}$

1. Check the form: As $x\to \infty$, the base $\frac{x+3}{x-1}\to 1$. The exponent $x+2\to \infty$. This is a $1^{\infty }$ form.
2. Identify $f(x)$ and $g(x)$:
   • $f(x)=\frac{x+3}{x-1}$
   • $g(x)=x+2$
3. Apply the formula: We need to calculate $L=e^K$ where $K={\lim }_{x\to \infty }g(x)[f(x)-1]$.
   $K={\lim }_{x\to \infty }(x+2)\left[\frac{x+3}{x-1}-1\right]$
4. Simplify the expression in the bracket:
   $\frac{x+3}{x-1}-1=\frac{x+3-(x-1)}{x-1}=\frac{4}{x-1}$
5. Evaluate the limit for $K$:
   $K={\lim }_{x\to \infty }(x+2)\left[\frac{4}{x-1}\right]={\lim }_{x\to \infty }\frac{4x+8}{x-1}$
   This is a limit at infinity of a rational function. Since the degrees of the numerator and denominator are the same, the limit is the ratio of the leading coefficients.
   $K=\frac{4}{1}=4$
6. Final Answer: The original limit is $e^K=e^4$.

Tackling $0^0$ and ${\infty }^0$

For these forms, the $f(x)-1$ shortcut does not apply. We must return to the general logarithmic method. The goal is always to find the limit of the exponent, $K={\lim }_{x\to a}[g(x)\cdot \ln (f(x))]$.

• For $0^0$, we have $g(x)\to 0$ and $f(x)\to 0^{+}$, so $\ln (f(x))\to -\infty$. The exponent is a $0\cdot (-\infty )$ form.
• For ${\infty }^0$, we have $g(x)\to 0$ and $f(x)\to \infty$, so $\ln (f(x))\to \infty$. The exponent is a $0\cdot \infty$ form.

In both cases, we must transform the product into a fraction and use our standard limit toolkit. Often, this requires the more advanced L'Hôpital's Rule, which we will cover later. However, some problems can be solved by clever algebraic manipulation and identifying dominant terms, a key skill in physics and engineering.

Let's try a ${\infty }^0$ problem that can be solved with our current tools.
Evaluate ${\lim }_{x\to \infty }(e^x+x{)}^{1/x}$.

1. Check the form: The base $(e^x+x)\to \infty$ and the exponent $1/x\to 0$. This is ${\infty }^0$.
2. Apply the general logarithmic method: Let the limit be $L$. We need to find $K={\lim }_{x\to \infty }\ln \left((e^x+x{)}^{1/x}\right)$.
   $K={\lim }_{x\to \infty }\frac{1}{x}\ln (e^x+x)$
3. Simplify the logarithm: This is the key step. When $x$ is very large, $e^x$ is much, much larger than $x$. We can factor out this dominant term from inside the logarithm.
   $\ln (e^x+x)=\ln \left(e^x\left(1+\frac{x}{e^x}\right)\right)$
   Using the log property $\ln (ab)=\ln (a)+\ln (b)$:
   $\ln (e^x+x)=\ln (e^x)+\ln \left(1+\frac{x}{e^x}\right)=x+\ln \left(1+\frac{x}{e^x}\right)$
4. Substitute back and evaluate the limit for $K$:
   $K={\lim }_{x\to \infty }\frac{1}{x}\left[x+\ln \left(1+\frac{x}{e^x}\right)\right]={\lim }_{x\to \infty }\left[1+\frac{\ln (1+x/e^x)}{x}\right]$
   As $x\to \infty$, the term $x/e^x\to 0$ because exponential growth dominates polynomial growth. Since its argument is approaching 0, the numerator $\ln (1+x/e^x)$ also approaches 0. The denominator $x$ approaches $\infty$. Therefore, the entire fraction goes to 0.
   $K=1+0=1$
5. Final Answer: The original limit is $e^K=e^1=e$.

Conclusion

In this lesson, we expanded our arsenal to handle the exponential indeterminate forms $1^{\infty }$, $0^0$, and ${\infty }^0$. We saw that they represent a conflict between the base and the exponent, requiring a more sophisticated approach than direct substitution.

Key Takeaways:

• Exponential forms $1^{\infty }$, $0^0$, and ${\infty }^0$ are indeterminate because the final value depends on the relative rates at which the base and exponent approach their limits.
• The universal strategy is logarithmic transformation: To find $\lim f(x{)}^{g(x)}$, we evaluate $K=\lim [g(x)\ln (f(x))]$ and the answer is $e^K$.
• For the common $1^{\infty }$ form, this simplifies to the powerful shortcut formula: $\lim f(x{)}^{g(x)}=e^{\lim g(x)[f(x)-1]}$.
• For $0^0$ and ${\infty }^0$ forms, we use the general logarithmic method and often need to identify and factor out the "dominant term" to simplify the expression into a solvable form.

We've now completed our foundational study of limits. You have the tools to handle almost any limit problem thrown at you. In the next module, we will officially begin our journey into differentiation. We will start by defining the derivative from first principles, which, as you'll see, is itself a limit of the form $\frac{0}{0}$, bringing our journey full circle.