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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives

Evaluate the following limits: $ \lim\limits_{x \to -2} \Large\frac{ \Large\frac{1}{x}+ \Large\frac{1}{2}}{ \normalsize x +2}$

$\begin{array}{1 1} \frac{-1}{4} \\ \frac{-1}{2} \\ \frac{1}{4} \\ \frac{1}{2} \end{array} $

1 Answer

$ \lim\limits_{x \to -2} \large\frac{ \Large\frac{1}{x}+ \Large\frac{1}{2}}{ x +2}$
At $ x = -2$ the value of the given function takes the form $ \large\frac{0}{0}.$
Now, $ \lim\limits_{x \to -2} \large\frac{ \Large\frac{1}{x}+ \Large\frac{1}{2}}{ x +2}$ $ = \lim\limits_{x \to -2} \large\frac{\bigg( \Large\frac{2+x}{2x} \bigg) }{x+2}$
$ = \lim\limits_{x \to -2} \large\frac{ 1}{2x}$
$ \large\frac{1}{2(-2)}$$ = \large\frac{-1}{4}$

 

answered Apr 3, 2014 by thanvigandhi_1
 

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