Question

Examine the continuity of
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 1 over denominator straight x plus 1 end fraction comma space straight x not equal to negative 1 end cell row cell space space minus 2 space space space space space comma space straight x equals negative 1 end cell end table close$

Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus 1 over denominator straight x plus 1 end fraction comma space straight x not equal to negative 1 end cell row cell space space minus 2 space space space space space comma space straight x equals negative 1 end cell end table close space space space space space Lt with straight x rightwards arrow negative 1 below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow negative 1 below fraction numerator straight x squared minus 1 over denominator straight x plus 1 end fraction space space equals stack space Lt with straight x rightwards arrow negative 1 below fraction numerator left parenthesis straight x plus 1 right parenthesis left parenthesis straight x minus 1 right parenthesis over denominator straight x plus 1 end fraction equals space Lt with straight x rightwards arrow negative 1 below left parenthesis straight x minus 1 right parenthesis space space space space space space space space space space space space space space space space space space space space space equals negative 1 minus 1 equals negative 2 Also space straight f left parenthesis negative 1 right parenthesis equals negative 2 therefore Lt with straight x rightwards arrow negative 1 below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis negative 1 right parenthesis

∴ f(x) is continuous at x = – 1

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