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Chapter 4: Problem 50

Evaluate the following limits. $$\lim _{x \rightarrow 0^{+}}(\sin x) \sqrt{\frac{1-x}{x}}$$

Short Answer

Expert verified
Expression: \(\lim _{x \rightarrow 0^{+}}(\sin x) \left(\frac{1-x}{x}\right)^{\frac{1}{2}}\) Answer: The limit of the expression as x approaches 0 from the positive side is 0.

Step by step solution

01

Rewrite the expression

It's more convenient to rewrite the expressions with a fractional exponent: $$\lim _{x \rightarrow 0^{+}}(\sin x) \left(\frac{1-x}{x}\right)^{\frac{1}{2}}$$
02

Evaluate the limit of the sine function

Since sine function is continuous at x = 0 and \(\sin(0) = 0\), we have: $$\lim _{x \rightarrow 0^{+}}\sin x = 0$$
03

Rewrite the second part as a single fraction and apply L'H么pital's rule

As L'H么pital's rule works for tending to infinity, let's rewrite \(u=\frac{x}{1-x}\). So when \(x\to 0^{+}\), \(u \to 0^{+}\). Along with converting to \(u\), we cleanup and apply L'H么pital's rule to find the limit: $$\lim_{x\to0^{+}} \left(\frac{1-x}{x}\right)^{\frac{1}{2}} = \lim_{u\to0^{+}} (u)^{\frac{1}{2}} = \lim_{u\to0^{+}} \dfrac{1}{\left(\frac{2}{u^{\frac{-1}{2}}}\right)}$$ By applying L'H么pital's rule, we obtain: $$\lim_{u\to0^{+}} \dfrac{1}{\left(\frac{2}{u^{\frac{-1}{2}}}\right)} = \lim_{u\to0^{+}} \dfrac{u^{\frac{1}{2}}}{2}$$
04

Evaluate the limit of the second part

By plugging \(u = 0\) into the expression, we get: $$\lim_{u\to0^{+}} \dfrac{u^{\frac{1}{2}}}{2} = \dfrac{0^{\frac{1}{2}}}{2} = 0$$
05

Combine the limits

Since both parts of the expression have limits that equal to 0, the limit of the entire expression is: $$\lim_{x\rightarrow 0^{+}}(\sin x) \left(\frac{1-x}{x}\right)^{\frac{1}{2}} = 0 \cdot 0 = 0$$

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