#lim_(x to 0) sin^5(x)/ (x^2+x) sin(x^4)# how should i solve it?

Question

im supposed to use the #lim_(x to 0) sin(x) / x = 1# or #lim_(u(x) to 0) sin(u(x)) /(u(x)) = 1# formulas

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Answer 1 · 2018-01-28T17:14:15

#0#

#lim_(x->0)sin^5(x)/(x^2+x)sin(x^4)#

Take the #sin^5(x)# and split it up like so:

#=lim_(x->0)sin(x)/(x^2+x)sin^4(x)sin(x^4)#

#=lim_(x->0)sin(x)/(x^2+x)lim_(x->0)sin^4(x)sin(x^4)#

So, evaluating these limits separately :

#lim_(x->0)sin^4(x)sin(x^4)=0times0#

and

#lim_(x->0)sin(x)/(x^2+x)=(lim_(x->0)sin(x))/(lim_(x->0)(x^2+x))#

For the denominator, the #x^2# term will go to #0# much faster than the #x# term so it will vanish leaving us with:

#=(lim_(x->0)sin(x))/(lim_(x->0)(x^2+x))=(lim_(x->0)sin(x))/(lim_(x->0)x)#

#=lim_(x->0)(sin(x))/(x)=1#

Hence:

#lim_(x->0)sin^5(x)/(x^2+x)sin(x^4)#

#=lim_(x->0)sin(x)/(x^2+x)lim_(x->0)sin^4(x)sin(x^4)#

#1times0times0=0#