Evaluating Limits Involving Integrals: A Taylor Series Approach

In this article, we will explore how to solve a specific limit problem that involves both integrals and Taylor series. We will walk through the detailed steps and use the well-known Taylor series expansion of the exponential function to achieve the solution. The limit in question is as follows:

$$L lim_{x to 0} frac{1}{x^5} int_{0}^{x} e^{-t^2} dt - frac{1}{x^4} frac{1}{3x^2}$$

### Step 1: Understanding the Problem The given limit consists of two main components: the integral of the exponential function and a rational term. To solve this, we will first expand the integral using the Taylor series of the exponential function. ### Step 2: Expanding the Integral The Taylor series expansion of the exponential function (e^x) is given by:

$$e^x sum_{r0}^{infty} frac{x^r}{r!}$$

Using this, we can express (e^{-t^2}) as a series:

$$e^{-t^2} sum_{r0}^{infty} frac{(-1)^r t^{2r}}{r!}$$

The integral (int_{0}^{x} e^{-t^2} dt) can be written using the series expansion as follows:

$$int_{0}^{x} e^{-t^2} dt int_{0}^{x} left(1 - frac{t^2}{1!} - frac{t^4}{2!} - frac{t^6}{3!} - ldotsright) dt$$

Evaluating the integral term by term, we get:

$$int_{0}^{x} e^{-t^2} dt left[x - frac{x^3}{3 cdot 1!} - frac{x^5}{5 cdot 2!} - ldotsright]$$

Simplifying further, the integral becomes:

$$int_{0}^{x} e^{-t^2} dt x - frac{x^3}{3} - frac{x^5}{10} - frac{x^7}{42} - ldots$$

### Step 3: Solving the Limit Substitute the integral back into the limit expression. We get:

$$L lim_{x to 0} frac{1}{x^5} left(x - frac{x^3}{3} - frac{x^5}{10} - frac{x^7}{42} - ldotsright) - frac{1}{x^4} frac{1}{3x^2}$$

Simplify the expression inside the limit:

$$L lim_{x to 0} left(frac{1}{x^4} - frac{1}{x^4} cdot frac{1}{3} - frac{x^2}{1^4} - frac{x^4}{42x^4} - ldots right) - frac{1}{x^4} frac{1}{3x^2}$$

This simplifies to:

$$L lim_{x to 0} left(frac{1}{x^4} - frac{1}{3x^2} - frac{1}{10} - frac{x^2}{42} - ldots right) - frac{1}{3x^2}$$

The terms that do not include (x) are isolated, and we are left with the expression:

$$L -frac{1}{10}$$

However, this is incorrect, as we need to re-evaluate the simplified limit expression. Let's use L'H?pital's rule for the correct solution. ### Step 4: Applying L'H?pital's Rule By L'H?pital's rule, we can rewrite the limit as follows:

$$L lim_{x to 0} frac{3 int_{0}^{x} e^{-t^2} dt - 3x^3}{3x^5}$$

Applying the Newton-Leibniz formula, we get the derivative of the integral as:

$$frac{d}{dx} int_{0}^{x} e^{-t^2} dt e^{-x^2}$$

Substituting this back, we get:

$$L lim_{x to 0} frac{3e^{-x^2} - 3x^2}{3x^5}$$

This still gives an indeterminate form (frac{0}{0}), so we apply L'H?pital's rule again:

$$L lim_{x to 0} frac{-6xe^{-x^2} - 6x}{15x^4} lim_{x to 0} frac{-6(e^{-x^2} - 1)}{15x^3}$$

Applying L'H?pital's rule one more time:

$$L lim_{x to 0} frac{12xe^{-x^2} 12x 12}{45x^2} lim_{x to 0} frac{12(xe^{-x^2} 1)}{45x^2} frac{12}{45} frac{1}{10}$$

### Conclusion Therefore, the limit (L) evaluates to:

$$L frac{1}{10}$$

This result shows the power of using Taylor series and L'H?pital's rule in solving complex limits involving integrals. Understanding these methods can greatly enhance problem-solving skills in advanced calculus and analysis.