Q. 18 鈭�-鈭瀟hat approaches (a)聽0聽(... [FREE SOLUTION]

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Chapter 1: Q. 18 (page 149)

$鈭� - 鈭�$ that approaches (a) $0$ (b) role="math" localid="1648201290597" $5,$ (c) $鈭� .$

Short Answer

Expert verified

$鈭� - 鈭�$ that approaches to (c) $鈭� .$

Step by step solution

Step 1. Given information

The given expression is $鈭� - 鈭�,$ we need to write $鈭� - 鈭�$ that approaches to (a) $0$ (b) $5$ (c) $鈭� .$

Step 2. The explanation for the correct option

Take an example to check the limits of the given form.

$\begin{array}{rcl} \lim_{x 鈫� 鈭�} \frac{5 (x^{2} 鈥� - 鈥� 1)}{x^{2} 鈥� - 鈥� x} & = & \lim_{x 鈫� 鈭�} \frac{5 \frac{d}{d x} (x^{2} 鈥� - 鈥� 1)}{\frac{d}{d x} (x^{2} 鈥� - 鈥� x)} [n u m a r t o r i s f o r m o f 鈭� - 鈭�] \\ = & \lim_{x 鈫� 鈭�} \frac{5 (2 x)}{(2 x 鈥� - 1)} \\ = & \lim_{x 鈫� 鈭�} \frac{5 (2 x)}{x (2 鈥� - \frac{1}{x})} \\ = & \lim_{x 鈫� 鈭�} \frac{10}{(2 鈥� - \frac{1}{x})} \\ = & \frac{10}{2} \\ = & 5 \end{array}$

And,

Like $\lim_{x 鈫� 鈭�} [\sqrt{x^{2} - x} - x] .$

localid="1654583934966" $= \lim_{x 鈫� 鈭�} [\sqrt{x^{2}} \sqrt{1 - \frac{1}{x}} - x] .$

$= \lim_{x 鈫� 鈭�} x [\sqrt{1 - \frac{1}{x}} - 1] .$

$= \lim_{x 鈫� 鈭�} \frac{\sqrt{1 - \frac{1}{x}} - 1}{\frac{1}{x}} .$

$= \lim_{x 鈫� 鈭�} \frac{(\frac{1}{2} \sqrt{1 - \frac{1}{x}}) (+ \frac{1}{x^{2}})}{- \frac{1}{x^{2}}} .$ [Applying L Hospital rule].

$= - \frac{1}{2} .$

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$鈭� - 鈭�$ that approaches (a) $0$ (b) role="math" localid="1648201290597" $5,$ (c) $鈭� .$

Short Answer

Step by step solution

Step 1. Given information

Step 2. The explanation for the correct option

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鈭�-鈭�that approaches (a)0(b) role="math" localid="1648201290597" 5,(c)鈭�.

Short Answer

Step by step solution

Step 1. Given information

Step 2. The explanation for the correct option

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

$鈭� - 鈭�$ that approaches (a) $0$ (b) role="math" localid="1648201290597" $5,$ (c) $鈭� .$