Among (i) `lim_(xtooo) sec^(-1)((x)/(sinx))" and "(ii) lim_(xtooo) sec^(-1)((sinx)/(x)).`A. `-(1)/(2sqrt(2))`B. `(1)/(2sqrt(2))`C. `(1)/(sqrt(2))`D. does not exist

Among (i) `lim_(xtooo) sec^(-1)((x)/(sinx))" and "(ii) lim_(xtooo) sec

1.	Among (i) `lim_(xtooo) sec^(-1)((x)/(sinx))" and "(ii) lim_(xtooo) sec^(-1)((sinx)/(x)).`A. `-(1)/(2sqrt(2))`B. `(1)/(2sqrt(2))`C. `(1)/(sqrt(2))`D. does not exist
Answer» Correct Answer - A `underset(xtooo)limsec^(-1)((x)/(sinx))=sec^(-1)((oo)/(sinoo))` `=sec^(-1)((oo)/("any value between "-1" to "1))` `=sec^(-1)(+-oo)=(pi)/(2)` `underset(xtooo)limsec^(-1)((sinx)/(x))=sec^(-1)((sinoo)/(oo))` `=sec^(-1)(("any value between "-1" to "1)/(oo))` `=sec^(-1)0` = not defined Hence, (i) exists but (ii) does not exist.

`lim_(xto oo) (1)/(n) {1+e^(1//n)+e^(2//n)+e^(3//n)+.....+e^((n-1)/(n))}` is equal toA. eB. `-e`C. `e-1`D. `1-e`
`lim_(xto oo) ((nsqrt(a)+nsqrt(b))/(2))^n,a,b,gt 0` equalsA. 1B. `sqrt(pq)`C. `pq`D. `(pq)/(2)`
If `lim_(xto oo){(x^2+1)/(x+1)-(ax+b)}to oo`, thenA. `a in (1,oo)`B. `a ne 1 , b in R`C. `a in (-oo,1)`D. none of these
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\(f\left(x\right)=\left[\left(\frac{min\left(t^2+4t+6\right)\sin \left(x\right)}{x}\right)\right]\)where [.] represents greatest integer function, then find \(\lim _{x\to 0}\left(f\left(x\right)\right)\)
`lim_(xrarr0) ((1-cos2x)^2)/(2xtanx-xtan2x)` , isA. 2B. `-(1)/(2)`C. `(1)/(2)`D. `-2`
The value of `lim_(xrarroo) ((x+3)/(x-1))^(x+1)` isA. eB. `e^2`C. `e^4`D. `1//e`
`lim_(xrarroo) ((x+2)/(x+1))^(x+3)` is equal toA. 1B. eC. `e^2`D. `e^3`
`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `(10)/(3)`B. `(3)/(10)`C. `(6)/(5)`D. `(5)/(6)`
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