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प्रश्न
Examine the function
f(x) `{:(= x^2 cos (1/x)",", "for" x ≠ 0),(= 0",", "for" x = 0):}`
for continuity and differentiability at x = 0
योग
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उत्तर
f'(0) = `lim_("h" -> 0) ("f"(0 + "h") - "f"(0))/"h"`
= `lim_("h" -> 0) ("f"("h") - 0)/"h"`
= `lim_("h" -> 0) ("h"^2 cos (1/"h"))/"h"`
= `lim_("h" -> 0) "h" cos(1/"h") ...[(because "h" -> 0),(therefore "h" ≠ 0)]`
= `[lim_("h" -> 0) "h"] xx [lim_("h" -> 0) cos(1/"h")]`
= [0] × [FInite number between – 1 and 1]
= 0
∴ f is differentiable at x = 0 and hence continuous at x = 0.
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