Find limx→0{tan(π4+x)1x}.
Consider the given function
y=limx→0⎧⎪ ⎪⎨⎪ ⎪⎩tan(π4+x)1x⎫⎪ ⎪⎬⎪ ⎪⎭
Taking log both side and we get,
logy=limx→0⎧⎪ ⎪⎨⎪ ⎪⎩logtan(π4+x)1x⎫⎪ ⎪⎬⎪ ⎪⎭
⇒logy=limx→0{1xlogtan(π4+x)}
⇒logy=limx→0logtan(π4+x)x
Its make 00from
Then,
Using L’ hospital rule and we get,
logy=limx→01tan(π4+x)×sec2(π4+x)1
logy=limx→01tan(π4+x)×sec2(π4+x)
logy=limx→0cos(π4+x)sin(π4+x)×1cos2(π4+x)
logy=limx→01sin(π4+x)cos(π4+x)
logy=limx→022sin(π4+x)cos(π4+x)
logy=limx→02sin2(π4+x)
logy=limx→02sin2(π4+x)
logy=limx→02csc2(π4+x)
logy=limx→02csc(π2+2x)
logy=limx→02sec2x
Taking limit and we get,
logy=2
logey=2
y=e2
Hence, this is the answer.