Question

Is the function f defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x comma space if space straight x less or equal than 1 end cell row cell 5 comma space if space straight x greater than 1 end cell end table close$
continous at x=0? At x=1? At x=2?

Solution

Here $straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x comma space if space straight x less or equal than 1 end cell row cell 5 comma space if space straight x greater than 1 end cell end table close$
At x=0
$Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 below left parenthesis straight x right parenthesis space space space space space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of minus right square bracket space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 0 minus straight h right parenthesis space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis negative straight h right parenthesis equals left parenthesis 0 right parenthesis equals 0 space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below left parenthesis straight x right parenthesis space space space left square bracket putx equals 0 plus straight h comma straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of plus right square bracket space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 0 plus straight h right parenthesis space space space space space space space space space space space space space equals 0 plus 0 equals 0 therefore Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals 0 therefore Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals 0 Also space straight f left parenthesis 0 right parenthesis equals value space of space straight x space at space straight x equals 0 space space space space space space space space space space space space equals 0 therefore straight f space is space continous space at space straight x equals 0. space space space space$
At x=1
$space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of minus below straight x space space space space left square bracket Put space straight x equals 1 minus straight h comma space straight h greater than straight o right square bracket comma space so space that space straight h rightwards arrow 0 space on space straight x rightwards arrow 1 to the power of minus right square bracket space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 1 minus straight h right parenthesis equals 1 minus 0 equals 1 space Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below left parenthesis 5 right parenthesis equals 5 therefore space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to Lt with straight x rightwards arrow 1 to the power of plus 0 end exponent below straight f left parenthesis straight x right parenthesis therefore space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis space does space not space exist therefore space straight f space is space discontinuous space at space straight x equals 1.$