Using the definition of continuity, explain why the beatiful piece-wise function f(x)={cos⁡(x),for x&lt;01,for x=01−x2,for x&gt;0 is continuous at x=0. Show work to support your explanation.

The given function is: f(x)={cos⁡(x),for x&lt;01,for x=01−x2,for x&gt;0 The function f(x) will be continuous at x=a when LHL=RHL=f(a) at x=0 LHL=limx→0−f(x)=limx→0−cos⁡x =limh→0cos⁡(0−h)=limh→0cos⁡(−h) =limh→0cos⁡h=cos⁡0=1 RHL=limx→0+f(x)=limx→0+(1−x2) =limh→0(1−(0+h)2)=limh→0(1−h2) =1−0 =RHL f(0)=1 LHL=RHL=f(0)=1 Hence f(x) is continuous at x=0

Answered: "Using the definition of continuity, explain why the..."

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UkusakazaL

Answered question

2021-10-30

Using the definition of continuity, explain why the beatiful piece-wise function $f (x) = {\begin{cases} \cos (x), & f o r x < 0 \\ 1, & f o r x = 0 \\ 1 - x^{2}, & f o r x > 0 \end{cases}$ is continuous at x=0. Show work to support your explanation.

Answer & Explanation

FieniChoonin

Skilled2021-10-31Added 102 answers

The given function is:
$f (x) = {\begin{cases} \cos (x), & f o r x < 0 \\ 1, & f o r x = 0 \\ 1 - x^{2}, & f o r x > 0 \end{cases}$
The function f(x) will be continuous at x=a when LHL=RHL=f(a)
at x=0
$L H L = lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} \cos x$
$= lim_{h \to 0} \cos (0 - h) = lim_{h \to 0} \cos (- h)$
$= lim_{h \to 0} \cos h = \cos 0 = 1$
$R H L = lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (1 - x^{2})$
$= lim_{h \to 0} (1 - {(0 + h)}^{2}) = lim_{h \to 0} (1 - h^{2})$
$= 1 - 0$
$= R H L$
$f (0) = 1$
LHL=RHL=f(0)=1
Hence f(x) is continuous at x=0

Jeffrey Jordon

Expert2022-06-24Added 2605 answers

Answer is given below (on video)

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