Question #d2302

1 Answer
Jim S
Nov 25, 2017

1) #~=2.1# , 2) #0# , 3) #1/2#

Explanation:

1) #lim_(xrarr4)(5x^4-4x^2-1)/(10-x-9x^3)# #=1215/570#

2) #lim_(xrarr+oo)x^2e^-x# #=# #lim_(xrarr+oo)x^2/e^x# #=#
#lim_(xrarr+oo)(2x)/e^x# #=# #2lim_(xrarr+oo)1/e^x# #=# #2*0# #=# #0# (Used rules De L'Hospital )

  • #lim_(xrarr+oo)e^x# #=# #+oo#

3) #lim_(xrarr0)(cosx-1)/x^2# #=# #lim_(xrarr0)((cosx-1)(cosx+1))/(x^2(cosx+1))# #=# #lim_(xrarr0)(cos^2x-1)/(x^2(cosx+1))# #=#
#lim_(xrarr0)-(1-cos^2x)/(x^2(cosx+1))# #=# #-lim_(xrarr0)(sin^2x)/(x^2(cosx+1)# #=# #-lim_(xrarr0)(sinx/x)^2*(1/(cosx+1))# #=# #1^2*1/2# #=1/2#

  • Note : #sin^2x+cos^2x=1# ,
    #lim_(xrarr0)sinx/x=1#
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