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Chapter 1: Problem 43

The value of $\lim _{x \rightarrow 0} \frac{(\tan (\\{x\\}-1)) \sin \\{x\\}}{\\{x\\}(\\{x\\}-1)}\(, where \)\\{x\\}$ denotes the fractional part function, is (A) is 1 (B) is tan 1 (C) is \(\sin 1\) (D) is non-existent

Short Answer

Expert verified
In conclusion, the limit as x approaches 0 of the given expression involving the fractional part function, trigonometric functions, and algebraic functions is 1. This evaluation involved using properties of limits, trigonometric functions, and the definition of the fractional part function. The key to solving this problem was recognizing and applying standard limits involving sin(x)/x and tangent functions.

Step by step solution

01

Rewrite the expression using properties of the fractional part function

The fractional part of x, denoted as \({x}\), can be defined as \(x-\lfloor x\rfloor\), where \(\lfloor x\rfloor\) is the greatest integer less than or equal to x. Since we are dealing with x approaching 0, we can assume x is a positive fraction, so \({x} = x\). We can rewrite the expression as: $$\lim _{x \rightarrow 0} \frac{(\tan (x-1)) \sin x}{x(x-1)}$$
02

Recognize the standard limit involving sin(x)/x

Recall the standard limit as x approaches 0: $$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ In our expression, we have the product of \(\tan (x-1)\) and \(\sin x\) in the numerator, and it will be beneficial to separate this product into two separate factors to make use of the standard limit.
03

Separate the factors and apply the standard limit

Rewrite the expression by separating the factors involving sin(x) and tan(x-1): $$\lim _{x \rightarrow 0} \frac{(\tan (x-1)) \sin x}{x(x-1)} = \lim _{x \rightarrow 0} \frac{\tan (x-1)} {x - 1} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x}$$ Now, apply the standard limit to the second factor: $$\lim _{x \rightarrow 0} \frac{\tan (x-1)} {x - 1} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x} = \lim _{x \rightarrow 0} \frac{\tan (x-1)} {x - 1} \cdot 1$$
04

Recognize another standard limit to apply

Another standard limit to recall is the limit of a tangent function, which is as follows: $$\lim_{x \rightarrow a} \frac{\tan x - \tan a}{x-a} = \sec^2 a$$ In our expression, we are considering the limit \(\lim_{x \rightarrow 1} \frac{\tan(x-1)}{x-1}\). Using the standard limit for the tangent function, we rewrite $$\lim _{x \rightarrow 1} \frac{\tan(x-1)} {x - 1} = \sec^2{0}$$
05

Evaluate the final value

Now, we can substitute the result from Step 4 into the expression and evaluate: $$\lim _{x \rightarrow 0} \frac{\tan (x-1)} {x - 1} \cdot 1 = \sec^2{0} \cdot 1$$ Since \(\sec^2{0} = 1^2 = 1\), we have: $$1 \cdot 1 = 1$$ The final value of the limit is 1, which corresponds to option (A).

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Most popular questions from this chapter

Consider the function \(f(x)=\left(\frac{a x+1}{b x+2}\right)^{x}\) where \(a^{2}+b^{2} \neq 0\) then \(\lim f(x)\) (A) exists for all values of \(a\) and \(b\) (B) is zero for \(\mathrm{a}<\mathrm{b}\) (C) is non existent for \(\mathrm{a}>\mathrm{b}\) (D) is e \({ }^{-\left(\frac{1}{a}\right)}\) or \(e^{-\left(\frac{1}{b}\right)}\) if \(a=b\)

Column-I (A) If \(\mathrm{f}(\mathrm{x})=|\mathrm{x}-\mathrm{a}|+|\mathrm{x}-10|+|\mathrm{x}-\mathrm{a}-10|\), where \(\mathrm{a} \in(0,10)\), then the minimum value of \(\mathrm{f}\) is (B) $\lim _{x \rightarrow 0} \frac{x(1-\cos 2 x)^{2}-a(\sin x-\tan x)^{2}}{\tan ^{5} x+a \sin ^{8} x}$ is equal to (C) $\lim _{n \rightarrow \infty} \frac{n^{a} \sin ^{2}(n !)}{n+1}, 00, a \neq 1$ is Column-II (P) 0 (Q) 1 (R) 4 (S) 10 (T) depends on a

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