Evaluating the Limit of (e^x - (1+x)) / x² as x approaches 0

We are asked to find the value of the limit:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}} \]

First, let's try substituting \(x = 0\) directly into the expression to see if we get an indeterminate form.

• Numerator: \(e^0 - (1 + 0) = 1 - 1 = 0\)
• Denominator: \(0^2 = 0\)

Since we get the indeterminate form \(\frac{0}{0}\), we can use L'Hôpital's Rule or a Taylor series expansion to evaluate the limit.

Using L'Hôpital's Rule

L'Hôpital's Rule states that if \(\mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'(x)}}{{g'(x)}}\), provided the latter limit exists.

Let \(f(x) = e^x - (1+x)\) and \(g(x) = x^2\).

Calculate the first derivatives:

• \(f'(x) = \frac{d}{dx}(e^x - 1 - x) = e^x - 1\)
• \(g'(x) = \frac{d}{dx}(x^2) = 2x\)

Now evaluate the new limit:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{f'(x)}}{{g'(x)}} = \mathop {\lim }\limits_{x \to 0} \frac{{e^x - 1}}{{2x}} \]

Substitute \(x = 0\) again:

• Numerator: \(e^0 - 1 = 1 - 1 = 0\)
• Denominator: \(2 \times 0 = 0\)

We still have the indeterminate form \(\frac{0}{0}\). So, we apply L'Hôpital's Rule again.

Calculate the second derivatives:

• \(f''(x) = \frac{d}{dx}(e^x - 1) = e^x\)
• \(g''(x) = \frac{d}{dx}(2x) = 2\)

Now evaluate the limit of the second derivatives:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{f''(x)}}{{g''(x)}} = \mathop {\lim }\limits_{x \to 0} \frac{{e^x}}{{2}} \]

Substitute \(x = 0\):

\[ \frac{{e^0}}{{2}} = \frac{{1}}{{2}} \]

So, the limit is \(\frac{1}{2}\).

Using Taylor Series Expansion

The Taylor series expansion for \(e^x\) around \(x = 0\) (Maclaurin series) is given by:

\[ e^x = 1 + x + \frac{{x^2}}{{2!}} + \frac{{x^3}}{{3!}} + \frac{{x^4}}{{4!}} + \dots \]

Substitute this expansion into the limit expression:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + x + \frac{{x^2}}{{2!}} + \frac{{x^3}}{{3!}} + \dots} \right) - \;\left( {1 + x} \right)}}{{{x^2}}} \]

Simplify the numerator:

\[ 1 + x + \frac{{x^2}}{2} + \frac{{x^3}}{6} + \dots - 1 - x = \frac{{x^2}}{2} + \frac{{x^3}}{6} + \dots \]

Substitute the simplified numerator back into the limit:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{x^2}}{2} + \frac{{x^3}}{6} + \dots}}{{{x^2}}} \]

Factor out \(x^2\) from the numerator:

\[ \mathop {\lim }\limits_{x \to 0} \frac{{x^2\left( {\frac{1}{2} + \frac{x}{6} + \dots} \right)}}{{{x^2}}} \]

Since \(x \to 0\), \(x \neq 0\), so we can cancel the \(x^2\) terms:

\[ \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{2} + \frac{x}{6} + \dots} \right) \]

Now substitute \(x = 0\):

\[ \frac{1}{2} + \frac{0}{6} + \dots = \frac{1}{2} \]

Both methods yield the same result.

Summary of Steps for Limit Evaluation

Evaluating limits often involves the following steps:

1. Substitute the limit value directly into the expression.
2. If a finite value is obtained, that is the limit.
3. If an indeterminate form (\(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), \(\infty^0\)) is obtained, apply techniques such as:
4. Algebraic simplification (factoring, rationalizing).
5. L'Hôpital's Rule (for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)).
6. Taylor series expansion (especially around 0).
7. Standard limits (e.g., \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\), \(\mathop {\lim }\limits_{x \to 0} \frac{{e^x - 1}}{x} = 1\)).

In this specific problem, the \(\frac{0}{0}\) indeterminate form required the use of L'Hôpital's Rule twice or the application of the Taylor series expansion for \(e^x\).

The final answer is \(\frac{1}{2}\).

Revision Table: Key Concepts for Limit Evaluation

Here's a quick review of concepts used in this problem:

• Limit: The value a function approaches as the input approaches some value.
• Indeterminate Form: An expression whose limit cannot be determined by simply evaluating the limits of its components (e.g., \(\frac{0}{0}\), \(\frac{\infty}{\infty}\)).
• L'Hôpital's Rule: A rule used to evaluate indeterminate forms by taking the derivatives of the numerator and denominator.
• Taylor Series: A representation of a function as an infinite sum of terms calculated from the function's derivatives at a single point. The Maclaurin series is a Taylor series centered at 0.

Additional Information: Exponential Function and its Series

The exponential function \(e^x\) is fundamental in calculus and has unique properties related to differentiation and series expansions. Its derivative is itself, \(\frac{d}{dx}(e^x) = e^x\). The fact that \(e^x\) can be represented by the infinite series \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\) makes it particularly useful for evaluating limits around \(x=0\), as seen in this problem. This series converges for all real values of \(x\).