Is the function f defined by f(x)= ll x, & if x 1 5, & if x>1 . continuous at x = 0 ? At x = 1 ? At x = 2 ?

It is given that f(x)= l x . if x 1 5, if x>5 . Case I : x = 0 We can see that f is defined at 0 and its value at 0 is 0. Left Hand Limit _ x 0^ - = _ h 0 f(0 - h) = _ h 0 - h = 0 Right Hand Limit _ x 0^ + = _ h 0 f(0 + h) = _ h 0 h = 0 Left Hand Limit = Right Hand Limit = f(0) Hence , f is continuous at x = 0. Case II : x = 1 We can see that f is defined at 1 and its value at 1 is 1. For x 1 f(x) = 5 therefore, Right Hand Limit _ x 1^ + f(x) = _ x 1^ + (5) = 5 Hence , f is not continuous at x = 1. Case III: x = 2 As, We can see that f is defined at 2 and its value at 2 is 5 Left Hand Limit: _ x 2^ - = _ h 0 f(2 - h) = _ h 0 5 = 5 Here f(2 - h) = 5, as h 0 2 - h 2 Right Hand Limit : _ x_ + = _ h + f(2 + h) = _ h 0 5 = 5 Left Hand Limit = Right Hand Limit = f(2) Here f(2 + h) = 5, as h h 0 2 - h 2 Hence , f is continuous at x = 2.

Is the function f defined by f(x)= ll x, & if x 1 5, & if x>1 . c…

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Question

Is the function $f$ defined by $f(x)=\left\{\begin{array}{ll} {x,} & {\text { if } x \leq 1} \\ {5,} & {\text { if } x>1} \end{array}\right.$ continuous at $x = 0$ ? At $x = 1$ ? At $x = 2$ ?

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Answer

It is given that $f(x)=\left\{\begin{array}{l} {x \text { . if } x \leq 1} \\ {5, \text { if } x>5} \end{array}\right.$
Case $I : x = 0$
We can see that $f$ is defined at $0$ and its value at $0$ is $0$.
Left Hand Limit $\mathop {\lim }\limits_{x \to {0^ - }} = \mathop {\lim }\limits_{h \to 0} f(0 - h) = \mathop {\lim }\limits_{h \to 0} - h = 0$
Right Hand Limit $\mathop {\lim }\limits_{x \to {0^ + }} = \mathop {\lim }\limits_{h \to 0} f(0 + h) = \mathop {\lim }\limits_{h \to 0} h = 0$
Left Hand Limit $=$ Right Hand Limit $= f(0)$
Hence $, f$ is continuous at $x = 0$.
Case $II : x = 1$
We can see that $f$ is defined at $1$ and its value at $1$ is $1$.
For $x < 1$
$f(x) = x$ Hence, Left Hand Limit:
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} x = 1$
For $x > 1 f(x) = 5$ therefore, Right Hand Limit
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (5) = 5$
Hence $, f$ is not continuous at $x = 1.$
Case $III: x = 2$
As, We can see that $f$ is defined at $2$ and its value at $2$ is $5$
Left Hand Limit:
$\mathop {\lim }\limits_{x \to {2^ - }} = \mathop {\lim }\limits_{h \to 0} f(2 - h)$ = $\mathop {\lim }\limits_{h \to 0} 5 = 5$
Here $f(2 - h) = 5, $as $h \longrightarrow 0 $
$\Rightarrow 2 - h \longrightarrow 2$ Right Hand Limit :
$\mathop {\lim }\limits_{{x_ + }} = \mathop {\lim }\limits_{h \to + \infty } f(2 + h) = \mathop {\lim }\limits_{h \to 0} 5 = 5$
Left Hand Limit $=$ Right Hand Limit $= f(2)$
Here $f(2 + h) = 5,$ as $h \ h \longrightarrow 0 $
$\Rightarrow 2 - h \longrightarrow 2$
Hence $, f$ is continuous at $x = 2.$