Math, asked by rinzinnamgyal6170, 7 months ago

सीमाओं के मान प्राप्त कीजिए : \lim_{x\rightarrow-2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}

Answers

Answered by amitnrw
0

-1/4     \lim_{x\rightarrow-2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2} = -\frac{1}{4}

Step-by-step explanation:

\lim_{x\rightarrow-2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}

x = -2  प्रयोग करने पर  

= (1/-2 + 1/2) /(-2 + 2)

= (-1/2 + 1/2)/0

= 0/0

परिभाषित नहीं

(1/x + 1/2)/(x + 2)

= {(2 + x)/2x}/(x + 2)

= (2 + x)/(2x(x + 2))

= (x + 2)/2x(x + 2)

= 1/2x

\lim_{x\rightarrow-2}\dfrac{1}{2x}

x = =2  प्रयोग करने पर  

= 1/2(-2)

= -1/4

\lim_{x\rightarrow-2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2} = -\frac{1}{4}

और पढ़ें

सीमाओं के मान प्राप्त कीजिए :  [tex]\lim_{x\rightarrow3}\dfrac{x^4 - 81

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