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  • What is the Right Hand limit of the function  <img alt="\lim _{x \rightarrow-2} \frac{|x|}{7 x^2-5}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crig

What is the Right Hand limit of the function  \lim _{x \rightarrow-2} \frac{|x|}{7 x^2-5} , where  \left |\\\ \right |  indicates the modulus function of the referred number?

Option: 1

\begin{aligned} & \frac{1}{6} \\ \end{aligned}


Option: 2

\begin{aligned} & -\frac{1}{16} \\ \end{aligned}


Option: 3

\begin{aligned} & -\frac{2}{27} \\ \end{aligned}


Option: 4

\begin{aligned} & \frac{2}{27} \end{aligned}


Answers (1)

best_answer

Note that the following essential points.

  • The Right Hand limit 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 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 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.

  The value of  |(-2+h)|  is  2-h  as h is positive and infinitely small.

The Right Hand limit of the function    \lim _{x \rightarrow-2} \frac{|x|}{7 x^2-5}   is

  \begin{aligned} & \lim _{x \rightarrow-2^{-}} \frac{|x|}{7 x^2-5} \\ & =\lim _{h \rightarrow 0} \frac{|(-2+h)|}{7(-2+h)^2-5} \\ & =\lim _{h \rightarrow 0} \frac{2-h}{7\left(4-4 h+h^2\right)-5} \\ & =\lim _{h \rightarrow 0} \frac{2-h}{32-32 h+7 h^2-5} \\ & =\lim _{h \rightarrow 0} \frac{2-h}{7 h^2-32 h+27} \\ & =\frac{2-0}{0-0+27} \\ & =\frac{2}{27} \end{aligned}

Posted by

HARSH KANKARIA

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