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प्रश्न
Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
उत्तर
Given that, f(x) = `{{:(x^2 sin 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
For differentiability we know that:
Lf'(c) = Rf'(c)
∴ Lf'(0) = `lim_("h" -> 0) ("f"(0 - "h") - "f"(0))/(-"h")`
= `lim_("h" -> 0) ((0 - "h")^2 sin 1/((0 - "h")) - 0)/(-"h")`
= `("h"^2 sin (- 1/"h"))/(-"h")`
= `"h"* sin (1/"h")`
= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`
= 0
Rf'(0) = `lim_("h" -> 0) ("f"(0 + "h") - "f"(0))/"h"`
= `lim_("h" -> 0) ((0 + "h")^2 sin (1/(0 + "h") - 0))/"h"`
= `lim_("h" -> 0) ("h"^2 sin (1/"h"))/"h"`
= `lim_("h" -> 0) "h" * sin (1/"h")`
= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`
= 0
So, Lf'(0) = Rf'(0) = 0
Hence, f(x) is differentiable at x = 0.
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