limits
posted by Reen .
lim as x> infinity sec^2(1/x)

As x > ∞ , 1/x > 0
So
lim sec^2(1/x) as x> infinity
= lim sec^2(x) as x> 0
= lim (1/cos^2(x) as x >0
= 1/1
= 1
Respond to this Question
Similar Questions

calculating infinity
can someone tell me wich one of the two answers for these two limits is correct? 
Maths (limits)
how do i evaluated these limits: lim x > 3 ((square root of x^2 +16) 5)/(x^2  3x) lim x> to infinity (2+ 100/x) 
Precal
Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x3x^2 / (2x+1)^2 x>  infinity lim 4n3 / 3n^2+2 n> infinity I did lim 2+6x3x^2 / (2x+1)^2 x>  infinity (2+6x3x²)/(4x²+4x+1) … 
Calc. Limits
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
calc
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
Calc Please Help
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
Calculus
Find the horizontal asymptote of f(x)=e^x  x lim x>infinity (e^x)x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0infinity =  infinity lim x> infinity (e^x)x= infinity 
Math
1. If 1/infinity = infinity or infinity ? 
Math
1. If 1/infinity = infinity or infinity ? 
Calculus
Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. …