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Question
Let f(x) be the continuous function such that f(x)=1−exx for x≠0 then.
Let f(x) be the continuous function such that f(x)=1−exx for x≠0 then.
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Solution
The correct option is C f′(0+)=f′(0−)=−12
f(x)=1−exx
Assuming f(0) to be 0,
f′(0+)=limh→0f(h)−f(0)h
⇒f′(0+)=limh→01−ehhh=limh→01−ehh2
Thus, f′(0+)=limh→0−eh2h=limh→0−eh2=−12
Now, f′(0−)=limh→0f(0)−f(−h)h
⇒f′(0−)=limh→0−1+e−h−hh=limh→0−1+e−h−h2
Thus, f′(0+)=limh→0−e−h−2h=limh→0e−h−2=−12 using the L'Hospital Rule.
f(x)=1−exx
Assuming f(0) to be 0,
f′(0+)=limh→0f(h)−f(0)h
⇒f′(0+)=limh→01−ehhh=limh→01−ehh2
Thus, f′(0+)=limh→0−eh2h=limh→0−eh2=−12
Now, f′(0−)=limh→0f(0)−f(−h)h
⇒f′(0−)=limh→0−1+e−h−hh=limh→0−1+e−h−h2
Thus, f′(0+)=limh→0−e−h−2h=limh→0e−h−2=−12 using the L'Hospital Rule.
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