Before reading this post, I would suggest checking the following thread: Is it possible lớn evaluate \$lim_x ightarrow 0(-1+cos x)^ an x\$? The answers provided by other users & my further đầu vào could be helpful.

I have the following limit that I should evaluate without using l"Hôpital:\$\$lim_x ightarrow 0 (-1+cos x)^ an x\$\$

Here"s what I"ve done:\$\$lim_x ightarrow 0 (-1+cos x)^ an x=lim_x ightarrow 0 e^ln((-1+cos x)^ an x)=lim_x ightarrow 0 e^ an x cdot ln(cos x-1)=lim_x ightarrow 0 e^sin x cdot fracln (cos x-1)cos x\$\$

but I"m stuck. Any help would be appreciated, thanks.

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calculus limits limits-without-lhopital
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edited Apr 13, 2017 at 12:19

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asked Dec 4, năm 2016 at 16:35

GlyceriusGlycerius
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\$egingroup\$
Claim: \$limlimits_x o 0 (cos(x)-1)^ an(x) = 1\$.Proof: We will need the half angle và double angle formulas:

Half angle formula: \$cos(x) -1 = -2sin^2(fracx2)\$

Double angle formula: \$sin(x) = 2sin(fracx2)cos(fracx2)\$

eginalign(cos(x)-1)^ an(x) &= extexpBig( an(x)log(-2sin^2(fracx2))Big) \ &= extexpBig( fracsin(x)cos(x)log(-2sin^2(fracx2))Big) \ &= extexpBig( frac2sin(fracx2)cos(fracx2) cos(x)2log(isqrt2sin(fracx2))Big) \ &= extexpigg(-i2sqrt2 underbraceigg(fraccos(fracx2)cos(x)igg)_=:g(x) underbraceBig(isqrt2sin(fracx2)Big)_=:f(x) log(isqrt2sin(fracx2))igg) \endalign

And here we have \$limlimits_x o 0 g(x) = 1\$ and \$limlimits_x o 0 f(x)log(f(x)) = 0\$, where the latter follows from \$limlimits_x o 0f(x) = 0\$ & \$limlimits_z o 0 zlog(z) = 0\$ for complex \$z\$.

Hence \$limlimits_x o 0 extexp(-i2sqrt2g(x)f(x)log(f(x))) = extexp(0) = 1\$