# To find $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{(\ln (\{x\}+|[x]|))}^{\{x\}}}$, where {} denotes fractional part function and [] denotes greatest integer function, is?

(a) 0

(b) 1

(c) ln 2

(d) ln $\dfrac{1}{2}$

Answer

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Hint: To solve the above problem, concepts of greatest integer function, fractional part function and basics of limits are required. Greatest integer function of any number x is the nearest integer to x which is less than or equal to x. Fractional part function basically gives the decimal part of any number (for 1.6, {1.6} = 0.6). We will use these properties to simplify the above function given inside the limits. In addition, we will also make use of the property, [x] = x – {x}.

Complete step by step answer:

Before beginning the problem, we try to understand the basic concepts which will be required to solve the problem. The greatest integer function ([x]) basically gives the nearest integer to x which is less than or equal to x. Thus, for example, for 1.9, we have [1.9] =1.

Further, [-1.2] = -2 and so on. In the case of the fractional part, this function gives the decimal part of any number. For example, in case of 1.6, {1.6} = 0.6). Now, we come back to the problem in hand, we have,

=$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{(\ln (\{x\}+|[x]|))}^{\{x\}}}$

Now, if x$\to {{0}^{-}}$, we can have h (for h>0) , x$\to $-h. Thus, we have

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+|[-h]|))}^{\{-h\}}}$

Since, h is close to zero and positive quantity, -h would be negative and close to zero (say

-0.001). Thus, [-h] = -1. Thus, substituting above,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+|-1|))}^{\{-h\}}}$

Now, |-1| = 1 (since, the absolute function of any number is the positive part of that number.

That is, |-2| =2 and so on.)

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+1))}^{\{-h\}}}$

Now, using, [h]=h-{h}, we have,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h-[-h])+1))}^{(-h-[-h])}}$

Since, [-h] = -1, we have,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h-(-1))+1))}^{(-h-(-1))}}$

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h+1)+1))}^{(-h+1)}}$

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (2-h))}^{(-h+1)}}$

Now, we can put h=0 in this above expression, we get,

= ln (2)

Hence, the correct answer is (c) ln (2).

Note: While solving questions related to one sided limits ($x\to {{0}^{+}}$ or $x\to {{0}^{-}}$), we substitute h in place of x (for $x\to {{0}^{+}}$) and (-h) [for $x\to {{0}^{-}}$] and then solve for h$\to $0. We try to solve the function inside the limits till we can successfully substitute h=0 (without getting an indeterminate form).

Complete step by step answer:

Before beginning the problem, we try to understand the basic concepts which will be required to solve the problem. The greatest integer function ([x]) basically gives the nearest integer to x which is less than or equal to x. Thus, for example, for 1.9, we have [1.9] =1.

Further, [-1.2] = -2 and so on. In the case of the fractional part, this function gives the decimal part of any number. For example, in case of 1.6, {1.6} = 0.6). Now, we come back to the problem in hand, we have,

=$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{(\ln (\{x\}+|[x]|))}^{\{x\}}}$

Now, if x$\to {{0}^{-}}$, we can have h (for h>0) , x$\to $-h. Thus, we have

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+|[-h]|))}^{\{-h\}}}$

Since, h is close to zero and positive quantity, -h would be negative and close to zero (say

-0.001). Thus, [-h] = -1. Thus, substituting above,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+|-1|))}^{\{-h\}}}$

Now, |-1| = 1 (since, the absolute function of any number is the positive part of that number.

That is, |-2| =2 and so on.)

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (\{-h\}+1))}^{\{-h\}}}$

Now, using, [h]=h-{h}, we have,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h-[-h])+1))}^{(-h-[-h])}}$

Since, [-h] = -1, we have,

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h-(-1))+1))}^{(-h-(-1))}}$

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln ((-h+1)+1))}^{(-h+1)}}$

= $\underset{h\to 0}{\mathop{\lim }}\,{{(\ln (2-h))}^{(-h+1)}}$

Now, we can put h=0 in this above expression, we get,

= ln (2)

Hence, the correct answer is (c) ln (2).

Note: While solving questions related to one sided limits ($x\to {{0}^{+}}$ or $x\to {{0}^{-}}$), we substitute h in place of x (for $x\to {{0}^{+}}$) and (-h) [for $x\to {{0}^{-}}$] and then solve for h$\to $0. We try to solve the function inside the limits till we can successfully substitute h=0 (without getting an indeterminate form).

Last updated date: 23rd Sep 2023

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