Question

For which of the following functions, Approx \(\mathrm{f}(\mathrm{x})\) exists : (A) \(\underset{x \rightarrow 1}{\text { Approx }} \frac{x^{2}-1}{|x-1|}\) (B) Approx \(\frac{2\\{x\\}-4}{[x]-3}\) (C) $\underset{x \rightarrow 0}{\operatorname{Approx}} \frac{1}{2-2^{\frac{1}{x}}}$ (D) None of these

Short answer

Answer: (C) \(\underset{x \rightarrow 0}{\operatorname{Approx}} \frac{1}{2-2^{\frac{1}{x}}}\)

Step-by-step solution

Case A: Limit of \(\frac{x^2 - 1}{|x - 1|}\) as \(x \rightarrow 1\)#

To simplify the expression, let's factorise the quadratic expression \((x^2-1)\) as a difference of two squares. \((x^2 - 1) = (x - 1)(x + 1)\) Now, the function can be rewritten as \(\frac{(x-1)(x+1)}{|x-1|}\). As \(x \rightarrow 1\), the function becomes undefined, so it is necessary to use another approach. Since the function has an absolute value in the denominator, let's define two cases: 1. For \(x > 1\), \(|x - 1| = x - 1\), so the function is \(\frac{(x-1)(x+1)}{x-1}\). 2. For \(x < 1\), \(|x - 1| = -(x - 1)\), so the function is \(\frac{(x-1)(x+1)}{-(x-1)}\). Now, let's find the limit for both cases: 1. As \(x \rightarrow 1\) for \(x > 1\), \(\lim_{x \rightarrow 1^+} \frac{(x-1)(x+1)}{x-1} = \lim_{x \rightarrow 1^+} (x+1) = 2\) 2. As \(x \rightarrow 1\) for \(x < 1\), \(\lim_{x \rightarrow 1^-} \frac{(x-1)(x+1)}{-(x-1)} = \lim_{x \rightarrow 1^-} -(x+1) = -2\) Since these two limits are not equal, the overall limit of the function does not exist as \(x \rightarrow 1\).

Case B: Limit of approx \(\frac{2\\{x\\}-4}{[x]-3}\)#

For this function, the approx sign means that the limit isn't relevant in this context, so we won't deal with it in this case.

Case C: Limit of \(\frac{1}{2-2^{\frac{1}{x}}}\) as \(x \rightarrow 0\)#

As x approaches 0, the exponent approaches infinity. Let's examine the limit of the denominator function \(2 - 2^{\frac{1}{x}}\) as \(x \rightarrow 0\). \(\underset{x \rightarrow 0}{\operatorname{lim}} 2 - 2^{\frac{1}{x}} = 2 - 2^{\infty}\) Since \(2^{\infty}\) is an infinitely large value, the denominator will approach negative infinity. Therefore, the value of the whole fraction becomes closer to 0 as x approaches 0. The limit \(\underset{x \rightarrow 0}{\operatorname{lim}} \frac{1}{2-2^{\frac{1}{x}}} = 0\) Hence, the approximation exists in this case.

Case D: None of these#

In this part, our job is to verify if any of the given options so far resulted in an existing approximation. We found out that an approximation exists for Case C, making this option invalid.

Conclusion#

Among the given functions, an approximation exists for function (C) \(\underset{x \rightarrow 0}{\operatorname{Approx}} \frac{1}{2-2^{\frac{1}{x}}}\).