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Which of the following is true for the limit of the function \lim _{x \rightarrow 0} \frac{|11 x|}{11 x} , where  \left | \, \, \right | indicates  the modulus function of the referred number?

Option: 1

LHL exist but RHL does not exist

 


Option: 2

RHL exist but LHL does not exist


Option: 3

The limit does not exist.

 


Option: 4

Both RHL and LHL exist and equals to 1


Answers (1)

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  • The Right Hand Limit (RHL) of the function f\left ( x \right ) at x=a is \lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h) .

  • The Left Hand Limit (LHL) of the function f\left ( x \right ) at  x=a  is\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h)  .

  • Here, h is positive and infinitely small.

  • The limit exists only when

  • The Modulus Function indicates the absolute value denoted mathematically as |p| , for any mathematical values. This function  |p|  gives the non-negative value of any negative value or the absolute value, and the non-positive value of any positive value or the absolute value.

As h is positive and infinitely small,

  • The value of |11(0-h)|=|0-11 h|=|-11 h|=11 h
  • The value of |11(0+h)|=|0+11 h|=|+11 h|=11 h 

The Left Hand limit of the function  \lim _{x \rightarrow 0^{}} \frac{|11 x|}{11 x}

\begin{aligned} & =\lim _{h \rightarrow 0} \frac{|1(0-h)|}{11(0-h)} \\ & =\lim _{h \rightarrow 0} \frac{11 h}{-11 h} \\ & =-1 \quad[\quad h \neq 0] \end{aligned}

The right Hand limit of the function  \lim _{x \rightarrow 0^{-}} \frac{|11 x|}{11 x}

RHF

   \begin{aligned} & =\lim _{h \rightarrow 0} \frac{|1(0+h)|}{11(0+h)} \\ & =\lim _{h \rightarrow 0} \frac{11 h}{11 h} \\ & =1 \quad[\quad h \neq 0] \end{aligned}

As L H L \neq R H L , the limit does not exist.

Posted by

HARSH KANKARIA

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