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  • For the limit of the function   <img alt="\lim _{x \rightarrow 8} \frac{\tan \{3-x\}}{\{x-2\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarr

For the limit of the function   \lim _{x \rightarrow 8} \frac{\tan \{3-x\}}{\{x-2\}} ,  where  \{\, \} indicates the fractional part function, which of the following is true?

 

Option: 1

LHL exists but RHL does not exist


Option: 2

RHL exists but LHL does not exist


Option: 3

Neither LHL nor RHL does exist

 


Option: 4

Both RHL and LHL exist and equals to \tan 1


Answers (1)

best_answer

Note that the following essential points.

  • 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 \lim _{x \rightarrow a+} f(x)=\lim _{x \rightarrow a-} f(x)

  • For the real number p that can be an integer or a fraction or a decimal,  [p]  is used to indicate the greatest integer function and \{p\} is used to indicate the fractional or the decimal part of p. Thus, mathematically, p=[p]+\{p\} \Rightarrow\{p\}=p-[p].

   The following is deduced which is used in the further calculations.

 \begin{aligned} &\cdot \{6+h\}=(6+h)-[6+h]=6+h-6=h \\ & \cdot \{-5-h\}=(-5-h)-[-5+h]=-5-h-(-5)=-5-h+6=-h \\ & \cdot \{6-h\}=(6-h)-[6-h]=6-h-5=1-h \\ &\cdot \{-5+h\}=(-5+h)-[-5+h]=-5+h-(-5)=-5-h+5=1-h \end{aligned}

The Left Hand limit of  \lim _{x \rightarrow 8} \frac{\tan \{3-x\}}{\{x-2\}}   is

\begin{aligned} & \lim _{x \rightarrow 8}\left(\frac{\tan \{3-x\}}{\{x-2\}}\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan \{3-(8-h)\}}{\{(8-h)-2\}}\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan \{-5+h\}}{\{6-h\}}\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan (-h)}{(1-h)}\right) \\ & =\left(\lim _{h \rightarrow 0} \frac{\tan (-h)}{(1-h)}\right) \\ & =\left(\frac{\tan 1}{1}\right) \\ & =\tan 1 \end{aligned}

The Right Hand limit of  \lim _{x \rightarrow 8} \frac{\tan \{3-x\}}{\{x-2\}}

\begin{aligned} & \lim _{x \rightarrow 8^{+}}\left(\frac{\tan \{3-x\}}{\{x-2\}}\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan \{3-(8+h)\}}{\{(8+h)-2\}}\right. \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan \{-5-h\}}{\{6+h\}}\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\tan (1-h)}{h}\right) \\ & =\left(\frac{\lim _{h \rightarrow 0} \tan (1-h)}{\lim _{h \rightarrow 0} h}\right) \\ & =\text { undefined } \end{aligned}

Thus, the LHL of     \lim _{x \rightarrow 8} \frac{\tan \{3-x\}}{\{x-2\}}   exists but its RHL does not exist.

 

 

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