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…

Download App

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?

✓

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "1 Marks" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Continuity and Differentiability — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions