Show that the function f(x) defined by f ( x )= ll x x + x, & x>0 2, & x=0 is continuous at x =0 . 4(1-1-x) x , & x<0 .

To show that the given function is continuous at x = 0 we show that ( LHL )_ x=0 =( RHL )_ x=0 =f(0) (i) Here, we have f ( x )= ll x x + x, x;0 2, x=0 4(1-1-x) x , x<0 . Now , LHL = _ x 0^ - f(x)= _ x 0^ - 4(1-1-x) x = _ h 0 4[1-1-(0-h) 0-h = _ h 0 4[1-1+h] -h = _ h 0 4[1-1+h] -h 1+1+h 1+1+h = _ h 0 4 [(1)^2-(1+h)^2 ] -h[1+1+h] = _ h 0 4[1-(1+h)] -h[1+1+h] = _ h 0 -h 4 -h[1+1+h] = _ h 0 4 1+1+h =4 1+1 =4 2 =2 and RHL= _ x 0^ + f(x)= _ x 0^ + ( x x + x ) RHL = _ h 0 ( h h + h ) = _ h 0 h h + _ h 0 h =1+ 0 =1+1 =2 Also, given that x=0, f(x)=2 f(0)=2 Since, (LHL )_ x=0 =( RHL )_ x=0 =f(0)=2 Therefore ,f ( x ) is continuous at x =0.

Show that the function f(x) defined by f ( x )= ll x x + x, & x>0…

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Question

Show that the function $f(x)$ defined by $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, & x>0 \\ 2, & x=0 \text { is continuous at } x =0 . \\ \frac{4(1-\sqrt{1-x})}{x}, & x<0\end{array}\right.$

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Answer

To show that the given function is continuous at $x = 0$ we show that
$( \text{LHL} )_{x=0}=( \text{RHL} )_{x=0}=f(0) \ldots (i)$
Here, we have $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, x;0 \\ 2, x=0 \\ \frac{4(1-\sqrt{1-x})}{x}, x<0\end{array}\right.$
Now $, \text{LHL} =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{4(1-\sqrt{1-x})}{x}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1-(0-h)}}{0-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h}]}{-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h]}}{-h} \times \frac{1+\sqrt{1+h}}{1+\sqrt{1+h}}$
$=\lim _{h \rightarrow 0} \frac{4\left[(1)^2-(\sqrt{1+h})^2\right]}{-h[1+\sqrt{1+h]}}$
$=\lim _{h \rightarrow 0} \frac{4[1-(1+h)]}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{-h \times 4}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{4}{1+\sqrt{1+h}}$
$=\frac{4}{1+\sqrt{1}}=\frac{4}{2}=2$
and $\text{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}+\cos x\right)$
$\Rightarrow \text { RHL }=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}+\cos h\right)$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h}+\lim _{h \rightarrow 0} \cos h$
$=1+\cos 0$
$=1+1$
$=2$
Also, given that $x=0, f(x)=2 $
$\Rightarrow f(0)=2$
Since, $(\text{LHL} )_{x=0}=( \text{RHL} )_{x=0}=f(0)=2$
Therefore $,f ( x )$ is continuous at $x =0$.

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