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: limx12x=21=2
  2. Special Exponential and Logarithmic Limits:

    • Memorizing special exponential and logarithmic limits can simplify the evaluation process. For instance:
      • limx0(1+x)1x=e
      • limxln(x)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: limx0ex1x=limx0ex1=1
  4. L'Hôpital's Rule with Exponential and Logarithmic Functions:

    • In cases where the limit involves an indeterminate form (00 or ), applying L'Hôpital's Rule after transforming the expression to a form suitable for differentiation can be beneficial.
    • Example: limx0ln(1+x)x=limx0ddx(ln(1+x))ddx(x)=limx011+x1=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: limx0(1+3xe2x1)

Example: Evaluate the limit limx0e2x1x.

Solution:

  1. Direct Substitution:

    • If we substitute x=0 directly into the expression, we get e2010=e010=110.
    • This results in an indeterminate form 00, indicating that further steps are needed.
  2. Apply L'Hôpital's Rule:

    • Take the derivative of both the numerator and the denominator:
      • Derivative of e2x is 2e2x.
      • Derivative of x is 1.
    • The limit becomes limx02e2x1 after applying L'Hôpital's Rule.
  3. Re-evaluate the Limit:

    • Substituting x=0 into 2e2x1 gives 2e201=2e01=21=2.

Therefore, after applying L'Hôpital's Rule, the exponential limit limx0e2x1x evaluates to 2 as x approaches 0.