Find the limit lim x→0 sin 5x/x+x^3 ?

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Answer 1 · 2018-03-09T17:25:42

#lim_(x->0)(sin5x)/(x+x^3)=5#

So we have:

#lim_(x->0)(sin5x)/(x+x^3)#

Since plugging #0# in the place of #x# gives you #0/0#, you can use the L'Hopital's Rule.

The rule states that: #lim_(x-c)(f(x))/(g(x))=(f'(c))/(g'(c))# if #(f(c))/(g(c))# gives you an indeterminate form.

Therefore, #lim_(x->0)(sin5x)/(x+x^3)=lim_(x->0)(d/dx(sin5x))/(d/dx(x+x^3))#

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.
#d/dx(sinx)=cosx#

#d/dx(f(g(x)))=f'(g(x))*g'(x)#

We have:

#lim_(x->0)(d/dx(sin5x))/(d/dx(x+x^3))#

#=>lim_(x->0)(5cosx)/(1+3x^2)# Now plug #0# in the place of #x#.

#=>(5cos0)/(1+3*0^2)#

#=>(5*1)/(1+0)#

#=>(5)/(1)#

#=>5#

That's the answer!