The integer `n` for which `("lim")_(xvec0)((cosx-1)(cosx-ehatx)/(x^n)` is finite nonzero number is ________

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

We have,
`lim_(xto0) ((cos x-1)(cosx-e^x))/(x^n)`
`=lim_(xto0) ((1-cosx)/(x^2))((e^x-cosx)/(x^(n-2)))`
` =lim_(xto0) ((1-cos x)/(x^2))((e^(x)-1+1-cos x)/(x))(1)/(x^n-3)`
` =lim_(xto0) ((1-cos x)/(x^2)){(e^(x)-1)/(x)+(1-cos x)/(x)}(1)/(x^n-3)`
We know that
`lim_(xto0) (1-cos x)/(x^2)=(1)/(2), lim_(xto0) (e^x-1)/(x) =1 and lim_(xto0) (1-cos x)/(x)=0`.
Therefore, for the above limit to be non -zero, we must have
`n-3=0rArr n=3`
ALTER We know that
`e^x=1+(x)/(1!) +(x^2)/(2!) +(x^3)/(3!)+......and, cos x=1-(x^2)/(2!) +(x^4)/(4!) -(x^6)/(6!) +.....`
` therefore lim_(xto0) ((cos x-n)(cos x-e^x))/(x^n)`
` =lim_(xto 0) (1)/(x^n)(-(x^2)/(2!)+(x^4)/(4!)-(x^6)/(6!)+.....) (-(x)/(1!)-(2x^x)/(2!)-(x^3)/(3!)+(x^4)/(4!))`
` =lim_(xto0) (1)/(x^(n-3)) {(-(1)/(2!)+(x^2)/(4!)-(x^4)/(6!)+.....) (-(1)/(1!)-(2x)/(2!)-(x^2)/(3!)+(x^3)/(4!)....)}`
For this limit to exist and be finite non-zero, we must have
`n-3=0rArr n=3`.