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Question

Find limx0{tan(π4+x)1x}.

A
e5
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B
e3
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C
e5
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D
e2
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Solution

The correct option is D e2

Consider the given function

y=limx0⎪ ⎪⎪ ⎪tan(π4+x)1x⎪ ⎪⎪ ⎪

Taking log both side and we get,

logy=limx0⎪ ⎪⎪ ⎪logtan(π4+x)1x⎪ ⎪⎪ ⎪

logy=limx0{1xlogtan(π4+x)}

logy=limx0logtan(π4+x)x

Its make 00from

Then,

Using L’ hospital rule and we get,

logy=limx01tan(π4+x)×sec2(π4+x)1

logy=limx01tan(π4+x)×sec2(π4+x)

logy=limx0cos(π4+x)sin(π4+x)×1cos2(π4+x)

logy=limx01sin(π4+x)cos(π4+x)

logy=limx022sin(π4+x)cos(π4+x)

logy=limx02sin2(π4+x)

logy=limx02sin2(π4+x)

logy=limx02csc2(π4+x)

logy=limx02csc(π2+2x)

logy=limx02sec2x

Taking limit and we get,

logy=2

logey=2

y=e2

Hence, this is the answer.


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