How to Solve the Limit Using Taylor Series and Logarithmic Transformation

In mathematics, limits are often encountered in various forms, and understanding how to solve them can be quite challenging. One such problem involves solving the limit

L lim_{x rightarrow 0} left( frac{x}{1-sin x} right)^{frac{1}{x^2}}

Step-by-Step Solution Using Taylor Series Expansion

To solve this limit, we start by simplifying the expression inside the limit. As x approaches 0, we can use Taylor series expansions for sin x:

sin x x - frac{x^3}{6} O(x^5)

From this, we can approximate:

1 - sin x approx 1 - x frac{x^3}{6} O(x^5)

For small x, this simplifies to:

1 - sin x approx 1 - x

Simplifying the Expression

We can now write:

frac{x}{1-sin x} approx frac{x}{1-x} 1

This shows that as x approaches 0, the expression inside the limit approaches 1. To find how fast this approaches 1, we can further analyze the first order term:

frac{x}{1-x} 1 frac{x - x}{1-x} 1

To get a more precise approximation, we find the first non-zero term in the Taylor expansion:

1-sin x approx 1-x implies frac{x}{1-sin x} approx frac{x}{1-x} 1 - O(x^2)

Logarithmic Transformation for Closer Analysis

To analyze the limit more closely, we can rewrite the limit in a logarithmic form:

ln L lim_{x to 0} frac{1}{x^2} ln left( frac{x}{1-sin x} right)

Next, we compute:

ln left( frac{x}{1-sin x} right) ln x - ln (1-sin x)

Using the Taylor series approximation for ln (1-u) when u is small:

ln x - ln (1 - sin x) approx x - (sin x - frac{sin^2 x}{2} O(sin^3 x))

Since sin x approx x - frac{x^3}{6}, we get:

ln (1 - sin x) approx x - frac{x^3}{6} - frac{x^2}{2} O(x^4)

Substituting this back, we find:

ln left( frac{x}{1-sin x} right) approx x - (x - frac{x^2}{2}) frac{x^2}{2} O(x^3)

Final Limit Calculation

Now, substituting this into our logarithmic limit:

ln L lim_{x to 0} frac{1}{x^2} left( frac{x^2}{2} O(x^3) right) lim_{x to 0} left( frac{1}{2} O(x) right) frac{1}{2}

Exponentiating gives us:

L e^{frac{1}{2}} sqrt{e}

Thus, the final result is:

lim_{x to 0} left( frac{x}{1-sin x} right)^{frac{1}{x^2}} sqrt{e}

Conclusion

This solution demonstrates the power of Taylor series and logarithmic transformation in solving complex limits. By carefully applying these techniques, we can find the exact value of the limit, which is crucial in many areas of mathematics and engineering. If you're interested in learning more about solving limits with these methods, consider exploring similar problems or consulting advanced calculus textbooks.