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Chapter 2: Problem 240

$$ \left.\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x^{2}}-\sqrt[4]{1-2 x}}{x+x^{2}} \text { (Ans. } \frac{1}{2}\right\\} $$

Short Answer

Expert verified
The limit of the given expression as x approaches 0 is \(\frac{1}{2}\).

Step by step solution

01

Evaluating the limit directly

First, let's try plugging in x = 0 directly into the given expression: \(\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x^{2}}-\sqrt[4]{1-2x}}{x+x^2}\) The expression becomes: \(\frac{\sqrt[3]{1 + 0} -\sqrt[4]{1 - 0}}{0 + 0^2} = \frac{1 - 1}{0}\) We can see that this results in an indeterminate form \(\frac{0}{0}\). Thus, we need to use a different technique to evaluate this limit, and we move on to the next step.
02

Applying L'H么pital's Rule

Since the expression is in the form of a fraction, we can apply L'H么pital's Rule. This means we need to differentiate both the numerator and the denominator. Let's first differentiate the numerator: \(\text{Let }\quad y =\sqrt[3]{1+x^2} - \sqrt[4]{1-2x}\) Applying the chain rule and derivative of both terms: \(\frac{dy}{dx} =\frac{1}{3}(1+x^2)^{-\frac{2}{3}}(2x)-\frac{1}{4}(1-2x)^{-\frac{3}{4}}(-2)\) Now, differentiate the denominator: \(\text{Let }\quad z = x + x^2\) \(\frac{dz}{dx} = 1+2x\) Now, applying L'H么pital's Rule for the given limit: \(\lim _{x \rightarrow 0} \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \lim_{x \rightarrow 0} \frac{(1/3)(1+x^2)^{-2/3}(2x)- (1/4)(1-2x)^{-3/4}(-2)}{1+2x}\)
03

Simplifying and Evaluating the Expression

Let us now simplify the expression obtained in step 2 and evaluate the limit for x approaching 0: \(\lim_{x \rightarrow 0} \frac{(1/3)(1+x^2)^{-2/3}(2x)- (1/4)(1-2x)^{-3/4}(-2)}{1+2x}\) Evaluating the limit, we obtain: \(\frac{(1/3)(1)^{-2/3}(2 \cdot 0) - (1/4)(1)^{-3/4}(-2)}{1 + 2 \cdot 0}\) \(\Rightarrow \frac{(0) - (-\frac{1}{2})}{1}\) Upon simplifying, we get \(\frac{1}{2}\) as the limit. Answer: Thus, the limit of the given expression as x approaches 0 is \(\frac{1}{2}\).

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