Question

please ans continuity class 12th​

Answer 1

First let us understand some terms given in that question;-

Fractional part function{x}:-It refers to the fraction ,which have the value between0 to 1

$0 < \: x > 1$

Step up function[x]:-It have the constant value under the given condition of x

for example,

for 0<x>1 it's value will be constant

Sgn(x)=It's value is fixed under given condition of x

for,x>0 ,(x)=1

for,x<0,(x)=-1

for,x=0,(x)=0

Solution:-It is given that,if we put x=0,in the given term, then terms will become zero

i.e f(0)=0

The options are focusing whteher the limit of f(x) is continuous or not

so,let us find this

As we know,for continue limit,the f(x) should satisfy this condition

limit(LHS)=limit(RHS)=f(0)=0

Limit of the given term when,

$x \: trends \: to \: {0}^{ - }$

that x which is trending to 0^-,will be between -1 and 0

i.ex>0

in that case

[x]=some constant value

{x}=some value between 0 and 1

sgnx=-1

putting these value we get

LHL=

$\frac{some \: finite \: value \times {e}^{ {0}^{2} } \times (number \: between \: 0 \: and \: 1) }{ {e}^{ \frac{1}{ {0}^{2} } } \times ( - 1)} \\ \frac{1}{0 } = \infty \: so \: {e}^{ \infty } = \infty \\ \frac{some \: finite \: value \: \times 1 \times (number \: between \: 0 \: and \: 1)}{ \infty } \\ = 0 \\ hence \:left \: hand \: limit \: is \: 0 \\ now \: consider\: the \: right \: in \: which \: x = > {0}^{ + } \\ in \: that \: case \\ every \: thing \: will \: be \: same \: except \: sig(x) \\ which \: will \: be \: equal \: to \: 1 \: \\ putting \: the \: value \: again \: we \: get \\ left \: hand \: limit = \frac{(some \: finite \: value \times {e}^{0} \times (number \: in \: condion \: 0 < \: x > \: 1) }{ {e}^{ \infty } \times 1} \\ = 0 \\ this \: shows \: that \:$LHL=RHS=f(0)=0

this is the required condition of continuous limit

so,

the limit will be continuous at x=0

{hope it helps}

Answer 2

as the function is a constant function.its always zero

I don't think this would be that easy ....is it correct?