सिद्ध कीजिए कि फलन f(x) = 5x - 3, x = 0, x = -3 तथा x = 5 पर संतत है।
Exercise-5.1-1
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यहाँ, f(x) = 5x - 3
x = 0 पर, $ \lim \limits_{x \rightarrow 0}$ f(x) = $\lim \limits_{x \rightarrow 0}$ (5x - 3) = 5 $\times$ 0 - 3 = 0 - 3 = - 3
तथा f(0) = 5 $\times$ 0 - 3 = - 3
$\therefore$ $\lim \limits_{x \rightarrow 0} f(x)$ = f(0)
अतः f(x), x = 0 पर सतत् है।
x = -3 पर, $ \lim \limits_{x \rightarrow-3}$ f(x) = $ \lim \limits_{x \rightarrow-3}(5 x-3)$ = 5 $\times$(- 3) - 3 = -15 - 3 = -18
तथा f(- 3) = 5x - 3 - 3 = - 18
$\therefore $ $ \lim \limits_{x \rightarrow-3}$ f(x) = f(- 3)
अतः f(x), x = - 3 पर सतत् है।
x = 5 पर,$ \lim \limits_{x \rightarrow 5} f(x)$ = $\lim \limits_{x \rightarrow 5}$ (5x - 3) = 5 $\times$ 5 - 3 = 25 - 3 = 22
f(5) = 5 $\times$ 5 - 3 = 25 - 3 = 22
$\lim \limits_{x \rightarrow 5} $ f(x) = f(5)
तथा f(5) = 5 $\times$ 5 - 3 = 25 - 3 = 22
$\therefore$ $\lim \limits_{x \rightarrow 5}$ f(x) = f(5)
अतः f(x), x = 5 पर सतत् है।
x = 0 पर, $ \lim \limits_{x \rightarrow 0}$ f(x) = $\lim \limits_{x \rightarrow 0}$ (5x - 3) = 5 $\times$ 0 - 3 = 0 - 3 = - 3
तथा f(0) = 5 $\times$ 0 - 3 = - 3
$\therefore$ $\lim \limits_{x \rightarrow 0} f(x)$ = f(0)
अतः f(x), x = 0 पर सतत् है।
x = -3 पर, $ \lim \limits_{x \rightarrow-3}$ f(x) = $ \lim \limits_{x \rightarrow-3}(5 x-3)$ = 5 $\times$(- 3) - 3 = -15 - 3 = -18
तथा f(- 3) = 5x - 3 - 3 = - 18
$\therefore $ $ \lim \limits_{x \rightarrow-3}$ f(x) = f(- 3)
अतः f(x), x = - 3 पर सतत् है।
x = 5 पर,$ \lim \limits_{x \rightarrow 5} f(x)$ = $\lim \limits_{x \rightarrow 5}$ (5x - 3) = 5 $\times$ 5 - 3 = 25 - 3 = 22
f(5) = 5 $\times$ 5 - 3 = 25 - 3 = 22
$\lim \limits_{x \rightarrow 5} $ f(x) = f(5)
तथा f(5) = 5 $\times$ 5 - 3 = 25 - 3 = 22
$\therefore$ $\lim \limits_{x \rightarrow 5}$ f(x) = f(5)
अतः f(x), x = 5 पर सतत् है।
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