%3Cp%3EProve%2C%20using%26nbsp%3Ban%26nbsp%3Bepsilon%26nbsp%3B-%26nbsp%3Bdelta%2

Question

%3Cp%3EProve%2C%20using%26nbsp%3Ban%26nbsp%3Bepsilon%26nbsp%3B-%26nbsp%3Bdelta%26nbsp%3Bargument%26nbsp%3Bthat%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%3Cspan%26nbsp%3Bclass%3D%22finalized_jax%22%26nbsp%3Bdata-jax%3D%22%5Clim_%7Bx%5Crightarrow%26nbsp%3B0%7Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Clbrack%26nbsp%3B%5Csin%26nbsp%3B(x)%26nbsp%3B%2B%26nbsp%3B2%5Ccos%26nbsp%3B(x)%5Crbrack%26nbsp%3B%3D%26nbsp%3B0%22%3E%26nbsp%3B%5C(%5Clim_%7Bx%5Crightarrow%26nbsp%3B0%7Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Clbrack%26nbsp%3B%5Csin%26nbsp%3B(x)%26nbsp%3B%2B%26nbsp%3B2%5Ccos%26nbsp%3B(x)%5Crbrack%26nbsp%3B%3D%26nbsp%3B0%5C)%26nbsp%3B%3C%2Fspan%3E%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%3C%2Fp%3E

Explanation / Answer

let E(epsilon)>0 be given

take d(delta) = (E/3)^3

=>

we know that

|sinx +2cosx| <3.....(1)

=>

for |x|<d we have

|x|<(E/3)^3

=>

|x|^(1/s) < E/3

multiplying this with (1)

we get

|x^(1/3)[sin(x)+2cos(x)] - 0| = |x|^(1/3)*|sinx +2cosx|< E/3*3 = E

=>

limit = 0

thus proved

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