Evaluate the following limit:
`lim_(nto oo)(sum_(r=1)^(n) sqrt(r)sum_(r=1)^(h)1/(sqrt(r)))/(sum_(r=1)^(n)r)`
Correct Answer - `8/3`
`lim_(n to oo) (sum_(r=1)^(n)sqrt(r) sum_(r=1)^(n)1/(sqrt(4)))/(sum_(r=1)^(n)r)`
`:.` Limit `=lim_(nto oo) (1/n sum_(r=1)^(n)sqrt(4/n)(1/nsum_(r=1)^(n)sqrt(n/r)))/(1/n sum_(r=1)^(n)r/n)`
`=(int_(0)^(1)sqrt(x)dxint_(0)^(1)(dx)/(sqrt(x)))/(int_(0)^(1)x dx) =8/3`
In each of the given limits function is defined and contin-uously exists in the NBD of x where we have to find the limit. So, we will find the limits...
2 Answers 1 views`(i) underset(xto)lim(1-x)^((1)/(x))=(underset(xto0)lim(1-x)^((1)/(-x)))^(-1)=e^(-1)` (ii) Since, `underset(xto1)log_(e)x=0, underset(xto1)lim(1+log_(e)x)^((1)/(log_(e)x))=1` `(iii)underset(xto0)lim (1+sinx)^((1)/(x))=underset(xto0)lim((1+sinx)^((1)/(sinx)))^((sinx)/(x))` `=(underset(xto0)lim(1+sinx)^((1)/(sinx)))^(underset(xto0)lim^((sinx)/(x)))` `=e^(1)=e`
2 Answers 1 viewsCorrect Answer - C `underset(xto0)lim(a^(sqrt(x))-a^(1//sqrt(x)))/(a^(sqrt(x))+a^(1//sqrt(x))),agt1` Put `x=r^(2)."Then"` `underset(t to0)lim(a^(t)-a^(1//t))/(a^(t)+a^(1//t))` or`underset(t to0)lim(a^(t-1//t)-1)/(a^(t-1//t)+1)=(a^(-oo)-1)/(a^(-oo)+1)=-1`
2 Answers 1 viewsCorrect Answer - C `underset(xto0)lim(f(x))/(sin^(2)x)=8` `implies" "underset(xto0)lim((f(x))/(x^(2)))/((sin^(2)x)/(x^(2)))=8` `implies" "underset(xto0)lim(f(x))/(x^(2))=8" "...(1)` Also, `underset(xto0)lim(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-2(1-cosx))/(x^(2))-(e^(x)-1)/(x)+x)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-4"sin"^(2)(x)/(2))/(x^(2))=-1+0)=lamda` `implies" "underset(xto0)lim(g(x))/(x^(2))=-2lamda" "...(2)` Now, `underset(xto0)lim(1+2f(x))^((1)/(g(x)))`(one power infinity form) `=e^(underset(xto0)lim(2f(x))/(g(x)))=e^(underset(xto0)lim(2(f(x)//x^(2)))/((g(x)//x^(2))))` `=e^(underset(xto0)lim(16)/(-2lamda))` `=e^(-(8)/(lamda))` `e^(-1)`(given) `:." "lamda=8` `underset(xto0)lim(1+f(x))^((1)/(2g(x)))=e^(underset(xto0)lim(f(x))/(2g(x)))`...
2 Answers 1 viewsCorrect Answer - A `underset(xto0)lim(f(x))/(sin^(2)x)=8` `implies" "underset(xto0)lim((f(x))/(x^(2)))/((sin^(2)x)/(x^(2)))=8` `implies" "underset(xto0)lim(f(x))/(x^(2))=8" "...(1)` Also, `underset(xto0)lim(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-2(1-cosx))/(x^(2))-(e^(x)-1)/(x)+x)=lamda` `implies" "underset(xto0)lim((g(x))/(x^(2)))/((-4"sin"^(2)(x)/(2))/(x^(2))=-1+0)=lamda` `implies" "underset(xto0)lim(g(x))/(x^(2))=-2lamda" "...(2)` Now, `underset(xto0)lim(1+2f(x))^((1)/(g(x)))`(one power infinity form) `=e^(underset(xto0)lim(2f(x))/(g(x)))=e^(underset(xto0)lim(2(f(x)//x^(2)))/((g(x)//x^(2))))` `=e^(underset(xto0)lim(16)/(-2lamda))` `=e^(-(8)/(lamda))` `e^(-1)`(given) `:." "lamda=8` `underset(xto0)lim(1+f(x))^((1)/(2g(x)))=e^(underset(xto0)lim(f(x))/(2g(x)))`...
2 Answers 1 viewsCorrect Answer - `(1)` `underset(xtooo)lim(f(x)+(3f(x)-1)/(f^(2)(x)))=3` or `(underset(xtooo)limf(x)+(3underset(xtooo)limf(x)-1)/((underset(xtooo)limf(x))^(2)))=3` or `(y+(3y-1)/(y^(2)))=3` or `y^(3)-3y^(2)+3y-1=0` or `(y-1)^(3)=0` or `y=1`
2 Answers 1 viewsCorrect Answer - D `f(x)=lim_(nto oo)( 1/n)(x+1/n)^(2)+(x+2/n)^(2)+………+(x+(n-1)/n)^(2)` `=int_(0)^(1)(x+y)^(2)dy` `=x^(2)+x+1/3` `:. f(x)=(x+1/2)^(2)+1/3-1/4` `:.f(x)|_("min")=1/12`
2 Answers 1 viewsCorrect Answer - A::C `L=lim_(n to oo) (n^(3)sum_(r=1)^(n)e^(r//n))/((n+1)^(m)sum_(r=1)^(n)r^(2m))` `=lim_(n to oo) (n^(3)sum_(r=1)^(n)e^(r//n) . 1/n)/((n+1)^(m)n^(2m)sum_(r=1)^(n)(r/n)^(2m) . 1/n)` `=lim_(n to oo) (n^(3))/((n+1)^(m)n^(2m)) . (lim_(nto oo) 1/n sum_(r=1)^(n)e^(r//n))/(lim_(nto oo) 1/n sum_(r=1)^(n)(r/n)^(2m))` `=lim_(nto oo) (n^(3))/((n^(3)+n^(2))m)...
2 Answers 1 viewsCorrect Answer - (i) `pi/2`, (ii) 2, (iii) `12`
2 Answers 1 viewsCorrect Answer - `((2sqrt(e^(pi)))/(e^(2)))^(m)`
2 Answers 1 views