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Question
Prove that the function f defined by
f(x) = `{{:(x/(|x| + 2x^2)",", x ≠ 0),("k", x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.
Solution
we have f(x) = `{{:(x/(|x| + 2x^2)",", x ≠ 0),("k", x = 0):}`
At x = 0
L.H.L. = `lim_(x ->0^+) ((0 - "h"))/(|0 - "h"| + 2(0 - "h")^2`
= `lim_("h" -> 0) (-"h")/("h" + 2"h"^2)`
= `lim_("h" -> 0) (-1)/(1 + 2"h")`
= – 1
R.H.L. = `lim_(x -> 0^+) x/(|x| + 2x^2)`
= `lim_("h" -> 0) (0 + "h")/(|0 + "h"| + 2(0 + "h")^2`
= `lim_("h" -> 0) "h"/("h" + 2"h"^2)`
= `lim_("h" -> 0) 1/(1 + 2"h")`
= 1
Since, L.H.L. ≠ R.H.L. for any value of k.
Hence, f(x) is discontinuous at x = 0 regardless the choice of k.
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