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Question Number 29460 by prof Abdo imad last updated on 08/Feb/18
find   lim_(x→0)     ((e^(√(1+sinx))   −e)/(tanx)).
findlimx0e1+sinxetanx.
Answered by Cheyboy last updated on 09/Feb/18
lim_(x→0) ((e^(√(1+sin x)) −e)/(tanx))=0
limx0e1+sinxetanx=0
Commented by prof Abdo imad last updated on 22/Feb/18
for x−→0  we have  1+sinx ∼ 1 +x  and  (√(1+x)) ∼ 1+(x/2)  and e^(1+(x/2)) ∼e(1+(x/2)) ⇒  e^(√(1+sinx))  −e ∼ ((ex)/2)  but  tanx∼x ⇒  lim_(x→0)    ((e^(√(1+sinx)) −e)/(tanx)) = (e/2) .
forx0wehave1+sinx1+xand1+x1+x2ande1+x2e(1+x2)e1+sinxeex2buttanxxlimx0e1+sinxetanx=e2.
Answered by sma3l2996 last updated on 09/Feb/18
sinx∼_0 x⇔1+sinx∼_0 1+x  (√(1+sinx))∼_0 (1+x)^(1/2)   (√(1+x))∼_0 1+(1/2)x  e^(√(1+sinx)) ∼_0 e^(1+(1/2)x) =e×e^(x/2)   e^(x/2) ∼_0 1+(x/4)  e^(√(1+sinx)) −e∼_0 e(1+(x/4))−e=(e/4)x  tanx∼_0 x  lim_(x→0) ((e^(√(1+sinx)) −e)/(tanx))∼lim_(x→0) (((e/4)x)/x)=(e/4)
sinx0x1+sinx01+x1+sinx0(1+x)1/21+x01+12xe1+sinx0e1+12x=e×ex2ex/201+x4e1+sinxe0e(1+x4)e=e4xtanx0xlimx0e1+sinxetanxlimx0e4xx=e4
Commented by prof Abdo imad last updated on 22/Feb/18
 for u ∈v(0)  e^u  ∼ 1+u so e^(x/2) ∼1+(x/2)  .
foruv(0)eu1+usoex21+x2.
Commented by sma3l2996 last updated on 22/Feb/18
Yeah, you′re right
Yeah,youreright

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