Prove that the function f defined by f x - eft begin matrix x-x-2 x- 2 -if xlneq 0 1k- - if - x -0remains discontinuous at x -0- regardless the choice of k-

We have, f(x)={x/|x|+2x^2 amp;if x≠ 0 k, amp; if x=0 . At x=0,, LHL=x→ 0^ limx/|x|+2x^2=h→ 0lim(0 H)/|0 H|+2(0 H)^2 =h→ 0lim h/h+2h^2=h→ 0lim h/h(1+2h) ...

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Standard XII

Mathematics

Real Valued Functions

Prove that th...

Question

Prove that the function f defined by $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}} & i f x \neq 0 k, & i f x = 0 \end{matrix}$
remains discontinuous at x=0, regardless the choice of k.

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Solution

We have, $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}} & i f x \neq 0 k, & i f x = 0 \end{matrix}$
At x=0,,
LHL= $l i m x \to 0^{-} \frac{x}{| x | + 2 x^{2}} = l i m h \to 0 \frac{(0 - H)}{| 0 - H | + 2 (0 - H)^{2}} = l i m h \to 0 \frac{- h}{h + 2 h^{2}} = l i m h \to 0 \frac{- h}{h (1 + 2 h)} = - 1$
RHL= $l i m x \to 0^{+} \frac{x}{| x | + 2 x^{2}} = l i m h \to 0 \frac{0 + h}{| 0 + h | + 2 (0 + h)^{2}} = l i m h \to 0 \frac{h}{h + 2 h^{2}} = l i m h \to 0 \frac{h}{h (1 + 2 h)} = 1$
and f(0)=k
Since, LHL $\neq$ RHL for any value of k,
Hence, f(x) is discontinuous at x=0 regardless the choice of k.

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Prove that the function f defined by f(x)={x|x|+2x2if x≠0k,if x=0 remains discontinuous at x=0, regardless the choice of k.

Prove that the function f defined by $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}} & i f x \neq 0 k, & i f x = 0 \end{matrix}$
remains discontinuous at x=0, regardless the choice of k.