Maths > Limits > 7.0 Indeterminate Forms:

  Limits
    1.0 Introduction
    2.0 Definition of Limit - In a different form:
    3.0 Conditions for existence of Limit
    4.0 Some Standard Limits
    5.0 Algebra of limits
    6.0 Some Standard Methods of Evaluation of Limits:
    7.0 Indeterminate Forms:
    8.0 Sandwich Theorem / Squeeze Play Theorem:
    9.0 L'Hospital's Rule for evaluation of limits:

7.1 Limits of the form ${1^\infty }$
Let $f(x)$ and $g(x)$ be functions of $x$.

Let $\mathop {\lim }\limits_{x \to a} f(x) = 1$ and $\mathop {\lim }\limits_{x \to a} g(x) = \infty $.

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to a} f{(x)^{g(x)}}$ will be in the form of ${1^\infty }$.

Then

$\mathop {\lim }\limits_{x \to a} f{(x)^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to a} \left( {f(x) - 1} \right)g(x)}}$

Question 18.

Evaluate $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x}$.

Solution:

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x}$ is in the form of ${1^\infty }$.

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x} = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{2}{x} - 1} \right)x}} = {e^2}$

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