# Exponential and Logarithmic limits

### Evaluation of Exponential and Logarithmic Limits:

Objective: Exponential and logarithmic limits involve determining the value that a function containing exponential or logarithmic expressions approaches as the independent variable approaches a specific value or infinity.

Methods for Evaluating Exponential and Logarithmic Limits:

1. Direct Substitution:

• When substituting the value of $x$ into the function results in a defined expression, direct substitution can be used to evaluate the limit.
• Example: ${\mathrm{lim}}_{x\to 1}{2}^{x}={2}^{1}=2$
2. Special Exponential and Logarithmic Limits:

• Memorizing special exponential and logarithmic limits can simplify the evaluation process. For instance:
• ${\mathrm{lim}}_{x\to 0}\left(1+x{\right)}^{\frac{1}{x}}=e$
• ${\mathrm{lim}}_{x\to \mathrm{\infty }}\frac{\mathrm{ln}\left(x\right)}{x}=0$
3. Properties of Exponential and Logarithmic Functions:

• Utilizing properties of exponentials and logarithms, such as logarithm rules or exponential identities, can simplify expressions to facilitate limit evaluation.
• Example: ${\mathrm{lim}}_{x\to 0}\frac{{e}^{x}-1}{x}={\mathrm{lim}}_{x\to 0}\frac{{e}^{x}}{1}=1$
4. L'Hôpital's Rule with Exponential and Logarithmic Functions:

• In cases where the limit involves an indeterminate form ($\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$), applying L'Hôpital's Rule after transforming the expression to a form suitable for differentiation can be beneficial.
• Example: ${\mathrm{lim}}_{x\to 0}\frac{\mathrm{ln}\left(1+x\right)}{x}={\mathrm{lim}}_{x\to 0}\frac{\frac{d}{dx}\left(\mathrm{ln}\left(1+x\right)\right)}{\frac{d}{dx}\left(x\right)}={\mathrm{lim}}_{x\to 0}\frac{\frac{1}{1+x}}{1}=1$
5. Limits involving Exponential and Logarithmic Substitutions:

• For complex limits, applying substitutions involving exponential or logarithmic transformations can simplify expressions and enable evaluation.
• Example: ${\mathrm{lim}}_{x\to 0}\left(\frac{1+3x}{{e}^{2x}-1}\right)$

Example: Evaluate the limit ${\mathrm{lim}}_{x\to 0}\frac{{e}^{2x}-1}{x}$.

Solution:

1. Direct Substitution:

• If we substitute $x=0$ directly into the expression, we get $\frac{{e}^{2\cdot 0}-1}{0}=\frac{{e}^{0}-1}{0}=\frac{1-1}{0}$.
• This results in an indeterminate form $\frac{0}{0}$, indicating that further steps are needed.
2. Apply L'Hôpital's Rule:

• Take the derivative of both the numerator and the denominator:
• Derivative of ${e}^{2x}$ is $2{e}^{2x}$.
• Derivative of $x$ is $1$.
• The limit becomes ${\mathrm{lim}}_{x\to 0}\frac{2{e}^{2x}}{1}$ after applying L'Hôpital's Rule.
3. Re-evaluate the Limit:

• Substituting $x=0$ into $\frac{2{e}^{2x}}{1}$ gives $\frac{2{e}^{2\cdot 0}}{1}=\frac{2{e}^{0}}{1}=\frac{2}{1}=2$.

Therefore, after applying L'Hôpital's Rule, the exponential limit ${\mathrm{lim}}_{x\to 0}\frac{{e}^{2x}-1}{x}$ evaluates to $2$ as $x$ approaches 0.