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If f(x)= {x-|x|xif x≠02ifx=0 then show that limx→1f(x) does not exist. - Business Mathematics and Statistics

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प्रश्न

If f(x)= `{((x - |x|)/x if  x ≠ 0),(2 if x = 0):}` then show that `lim_(x->1)`f(x) does not exist.

योग

उत्तर

`"L"["f"(x)]_(x=0) = lim_(x->0^-) "f"(x) = lim_(h->0) "f"(0 - "h")`

`= lim_(h->0) "f"(-"h") = lim_(h->0) ((-"h") - |- "h"|)/(-"h")`

`= lim_(h->0) (- "h" - "h")/(- "h")`

`= lim_(h->0) (-2cancel("h"))/(-cancel(h))`

`= lim_(h->0) 2` = 2  ...[∵ |- h| = h]   ...(1)

`"R"["f"(x)]_(x=0) = lim_(x->0) "f"(x) = lim_(x->0) "f"(0 + "h")`

`= lim_(h->0)` f(h)

`= lim_(h->0) ("h" - |"h"|)/"h"`

`= lim_(h->0) ("h - h")/"h"`

`= lim_(h->0) 0/"h"` = 0   ...(2)

From (1) and (2),

`"L"["f"(x)]_(x=0) ne "R"["f"(x)]_(x=0)`

`therefore lim_(x-> ∞) "f"(x)` does not exist.

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Limits and Derivatives
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Differential Calculus - Miscellaneous Problems [पृष्ठ १२५]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
अध्याय 5 Differential Calculus
Miscellaneous Problems | Q 3 | पृष्ठ १२५
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