How do you evaluate the limit #cos((x^5+1)/(x^6+x^5+100))# as x approaches #-oo#?

Question

William Abdalla · Accepted Answer

The Reqd. Limit#=1#.

The Reqd. Limit #=lim_(xrarr-oo) cos {(x^5+1)/(x^6+x^5+100)}# #=lim_(xrarr-oo) cos {(x^5(1+1/x^5))/(x^6(1+1/x+100/x^6))}# #=lim_(xrarr-oo) cos {(1/x)((1+1/x^5)/(1+1/x+100/x^6))}# Since #cos# is continuous on #RR#, we have, The Reqd. Limit#=cos {lim_(xrarr-oo) (1/x)((1+1/x^5)/(1+1/x+100/x^6))}# #=cos {0*((1+0)/(1+0+0))}#. #=1#.

Isaiah Allen · Answer

undefined To evaluate the limit of cos((x^5+1)/(x^6+x^5+100)) as x approaches -oo, we can consider the highest power of x in the numerator and denominator. In this case, the highest power of x is x^6.

As x approaches -oo, x^6 approaches oo. Therefore, we can simplify the expression by dividing both the numerator and denominator by x^6.

After simplification, the expression becomes cos((1/x + 1/x^6 + 100/x^6)/(1 + 1/x + 100/x^6)).

As x approaches -oo, 1/x approaches 0 and 1/x^6 also approaches 0. Therefore, we can simplify the expression further by substituting these values.

After substitution, the expression becomes cos((0 + 0 + 0)/(1 + 0 + 0)).

Simplifying further, the expression becomes cos(0/1).

Since the cosine of 0 is equal to 1, the limit of cos((x^5+1)/(x^6+x^5+100)) as x approaches -oo is equal to 1.