`lim(x->0)(1/(x^5)int_0^xe^(-t^2)dt-1/(x^4)+1/(3x^2))` is equal to

Answer

To solve the limit

\[ \lim_{x \to 0} \left( \frac{1}{x^5} \int_0^x e^{-t^2} dt - \frac{1}{x^4} + \frac{1}{3x^2} \right), \]

we will follow these steps:

### Step 1: Expand \( e^{-t^2} \) using Taylor series
The Taylor series expansion for \( e^x \) is:

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

Substituting \( -t^2 \) for \( x \), we get:

\[ e^{-t^2} = 1 - t^2 + \frac{t^4}{2} - \frac{t^6}{6} + \cdots \]

### Step 2: Integrate \( e^{-t^2} \)
Now, we integrate \( e^{-t^2} \) from \( 0 \) to \( x \):

\[ \int_0^x e^{-t^2} dt = \int_0^x \left( 1 - t^2 + \frac{t^4}{2} - \frac{t^6}{6} + \cdots \right) dt \]

Calculating the integral term by term:

\[ = \left[ t - \frac{t^3}{3} + \frac{t^5}{10} - \frac{t^7}{42} + \cdots \right]_0^x \]

Evaluating at the limits gives:

\[ = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots \]

### Step 3: Substitute back into the limit expression
Now substitute this result back into the limit expression:

\[ \frac{1}{x^5} \left( x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots \right) \]

This simplifies to:

\[ = \frac{1}{x^5} \left( x - \frac{x^3}{3} + \frac{x^5}{10} \right) + \text{higher order terms} \]

Breaking it down:

\[ = \frac{1}{x^4} - \frac{1}{3x^2} + \frac{1}{10} - \frac{x^2}{42} + \cdots \]

### Step 4: Combine with the other terms in the limit
Now we substitute this into the limit:

\[ \lim_{x \to 0} \left( \left( \frac{1}{x^4} - \frac{1}{3x^2} + \frac{1}{10} - \frac{x^2}{42} + \cdots \right) - \frac{1}{x^4} + \frac{1}{3x^2} \right) \]

The \( \frac{1}{x^4} \) and \( -\frac{1}{x^4} \) cancel out, and the \( -\frac{1}{3x^2} \) and \( +\frac{1}{3x^2} \) also cancel out:

\[ = \lim_{x \to 0} \left( \frac{1}{10} - \frac{x^2}{42} + \cdots \right) \]

### Step 5: Evaluate the limit
As \( x \to 0 \), all terms involving \( x \) vanish:

\[ = \frac{1}{10} \]

Thus, the final answer is:

\[ \boxed{\frac{1}{10}} \]