Proving the Limit Without L'H?pital's Rule: ( lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} 1 )

Understanding how to find limits without relying on L'H?pital's rule can be a valuable skill in calculus. This article demonstrates how to prove the limit ( lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} 1 ) using standard theorems and properties of limits. We will explore several methods, including continuity, the Squeeze Theorem, and direct evaluation.

Using Continuity and Standard Limit Properties

The first step in evaluating the limit ( lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} ) is to understand that both the numerator and the denominator are sums of functions that are continuous at ( x 0 ).

Recall the following standard limits:

( lim_{x to 0} sin x 0 ) ( lim_{x to 0} sin^{-1} x 0 )

Since the functions ( sin x ) and ( sin^{-1} x ) are continuous at ( x 0 ), the function ( frac{1 sin x}{1 sin^{-1} x} ) is also continuous at ( x 0 ). Therefore, the limit can be found by simply evaluating the function at ( x 0 ).

[ lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} frac{1 sin 0}{1 sin^{-1} 0} frac{1 0}{1 0} 1 ]

Alternative Method Using the Squeeze Theorem

Another way to evaluate this limit is to use the Squeeze Theorem. Let's use the inequality ( sin x leq x ) for small ( x ). This inequality, combined with the fact that ( sin x ) and ( sin^{-1} x ) are close to each other for small values of ( x ), can help us simplify the limit.

First, we note that for ( x ) near 0:

( sin x approx x ) ( sin^{-1} x approx x )

Therefore, for small ( x ), the expression ( frac{1 sin x}{1 sin^{-1} x} ) is close to ( frac{1 x}{1 x} 1 ).

To formalize this with the Squeeze Theorem, consider the following inequalities:

[ 1 - |x| leq sin x leq x ]

For small ( x ), we can rewrite this as:

[ 1 - |x| leq sin x leq x ]

Adding 1 to all parts of the inequality:

[ 1 - |x| 1 leq 1 sin x leq 1 x ]

Dividing by 1 ( sin^{-1} x ) (which is also close to 1 for small ( x )):

[ frac{1 - |x| 1}{1 - |x| 1} leq frac{1 sin x}{1 sin^{-1} x} leq frac{1 x}{1 x} ]

Since both the lower and upper bounds simplify to 1, by the Squeeze Theorem:

[ lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} 1 ]

Using Limit Theorems Directly

Yet another approach involves using limit theorems directly. Specifically, if we have the limits ( lim_{x to 0} (1 sin x) 1 0 1 ) and ( lim_{x to 0} (1 sin^{-1} x) 1 0 1 ), we can use the theorem that states:

[ lim_{x to a} frac{f(x)}{g(x)} frac{lim_{x to a} f(x)}{lim_{x to a} g(x)} ]

provided that ( lim_{x to a} g(x) eq 0 ).

In this case:

[ lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} frac{lim_{x to 0} (1 sin x)}{lim_{x to 0} (1 sin^{-1} x)} frac{1}{1} 1 ]

Conclusion

By using continuity, the Squeeze Theorem, and standard limit theorems, we have shown that ( lim_{x to 0} frac{1 sin x}{1 sin^{-1} x} 1 ). This result showcases the power of these techniques in evaluating limits without resorting to L'H?pital's rule.

Keywords

limit, L'H?pital's rule, Squeeze Theorem, continuity, small values