Question

$\lim_{x\rightarrow1}f(x), \,ज्ञात कीजिए , जहाँ \,f(x) = \right \begin{cases}{^2x - 1, \,\,x \leq 1\atop -x^2 - 1, \, x \ \textgreater \ 1 \end{cases}$

Answer 1

Answer:

Step-by-step explanation:

$f(x) = \right \begin{cases}{x^2-1}, \,\,\,\,\,x \leq 1\atop -x^2-1, \, x \  \textgreater \  1 \end{cases}$

x > 1 के लिए  -

R.H.S.  =  $\lim_x_\rightarrow_1 f(x)=\lim_x_\rightarrow_1(-x^2-1)\\\\=-(-1)^2-1\\\\=-1-1\\\\=-2$

x < 1 के लिए  -

L.H.S.  = $\lim_x_\rightarrow_1 f(x)=\lim_x_\rightarrow_1(x^2-1)\\\\=(1)^2-1\\\\=1-1\\\\=0$

$\lim_x_\rightarrow_1{^-} f(x)\neq \lim_x_\rightarrow_1{^+} f(x)$

अतः  $\lim_{x\rightarrow1}f(x)$ का अस्तित्व नहीं है।