4. i 5. zadaca

[Vishykc]

Molim pomoć 4.b iz 5. zadaće: [tex]lim_{x\to-1}\frac{1+\sqrt[5]{x}}{1+\sqrt[3]{x}}[/tex]

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[quark]

Već je komentirano bilo (provjeri prvu stranicu ove teme za rješenje), pa ću citirati Zenona jer on koristi latex
Ovo ti je hint:

[anamarie]

5.zadaća
11.c)
meni ispada 9/4,a wolfram kaže 0,pa ne kužim gdje sam fulala..
uvela sam zamjenu t=-x

[Vishykc]

[quark]

[Zenon]

[quark]

[matematičarka]

Može li netko raspisati 11 pod c? Konstantno mi krivo ispada. 5. zadaća

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[Vishykc]

[Zenon]

Ima li neki bezbolniji način za ovo:
[dtex]\lim_{x\to0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[n]{\cos\beta x}}{x^2}, \ m,n\in\mathbb N, \ \alpha,\beta\in\mathbb R[/dtex]
[dtex]\lim_{x\to0}\frac{e^{\frac 1m\ln\left(\cos\alpha x\right)}-e^{\frac 1n\ln\left(\cos\beta x\right)}+1-1}{x^2}=\lim_{x\to0}\frac{\frac{e^{\frac 1m\ln\left(\cos\alpha x\right)}-1}{\frac 1m\ln\left(\cos\alpha x\right)}\cdot\frac{\ln\left(\cos\alpha x\right)}{m}-\frac{e^{\frac 1n\ln\left(\cos\beta x\right)}-1}{\frac 1n\ln\left(\cos\beta x\right)}\cdot\frac{\ln\left(\cos\beta x\right)}{n}}{x^2}=[/dtex]
[dtex]=\lim_{x\to0}\frac{\frac{e^{\frac 1m\ln\left(\cos\alpha x\right)}-1}{\frac 1m\ln\left(\cos\alpha x\right)}\cdot\frac{\ln\left(1+(\cos\alpha x-1)\right)}{\cos\alpha x-1}\cdot\frac{\cos\alpha x-1}{m}-\frac{e^{\frac 1n\ln\left(\cos\beta x\right)}-1}{\frac 1n\ln\left(\cos\beta x\right)}\cdot\frac{\ln\left(1+(\cos\beta x-1)\right)}{\cos\beta x-1}\cdot\frac{\cos\beta x-1}{n}}{x^2}=[/dtex]
[dtex]\lim_{x\to0}\left(-\frac{e^{\frac 1m\ln\left(\cos\alpha x\right)}-1}{\frac 1m\ln\left(\cos\alpha x\right)}\cdot\frac{\ln\left(1+(\cos\alpha x-1)\right)}{\cos\alpha x-1}\cdot\frac{1-\cos\alpha x}{(\alpha x)^2}\cdot\frac{\alpha^2}{m}+\frac{e^{\frac 1n\ln\left(\cos\beta x\right)}-1}{\frac 1n\ln\left(\cos\beta x\right)}\cdot\frac{\ln\left(1+(\cos\beta x-1)\right)}{\cos\beta x-1}\cdot\frac{1-\cos\beta x}{(\beta x)^2}\cdot\frac{\beta^2}{n}\right)=[/dtex]

[dtex]=-1\cdot 1\cdot\frac 12\cdot\frac{\alpha^2}{m}+1\cdot 1\cdot\frac 12\cdot\frac{\beta^2}{n}=\frac 12\left(\frac{\beta^2}{n}-\frac{\alpha^2}{m}\right)
[/dtex]


Unaprijed hvala!

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[tex]\heartsuit \ \mathcal{PMF-MO} \ \heartsuit[/tex]
[tex]\mathbb Z\Sigma\mathbb N\emptyset\mathbb N[/tex]

[gflegar]

Tamo negdje nakon 2. koraka rastavis na razliku limesa, izracunas jedan i napises da za drugi ide analogno, smanjis si posao na pola

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[Zenon]

[gflegar]

Bez brige, jasno je meni sto si ti mislio, ali mi se malo trolalo

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