Find lim x^4 - 8x/ x - 2 as x-->2 ??

Find lim (x^4 - 8x)/ (x - 2) as x-->2 ?

2 Answers
maganbhai P.
Jun 13, 2018

Answer for the question :
#lim_(xto2) (x^4-8x)/(x-2)=24#

Explanation:

We know that,

#color(red)((1)a^3-b^3=(a-b)(a^2+ab+b^2)#

Let,

#L=lim_(xto2) (x^4-8x)/(x-2)#

#=>L=lim_(xto2) (x(x^3-8))/(x-2)#

#=>L=lim_(xto2) (xcolor(red)((x^3-2^3)))/((x-2))...to color(red)(Apply(1)#

#=>L=lim_(xto2)(xcolor(red)(cancel((x-2))(x^2+2x+4)))/(cancel((x-2)))...to[x!=2]#

#=>L=lim_(xto2) [x(x^2+2x+4)]#

#=>L=[2(2^2+2(2)+4)]#

#=>L=[2(4+4+4)]#

#=>L=24#

Parabola
Jun 13, 2018

#lim_(x->2)(x^4-8x)/(x-2)=24#

Explanation:

We have: #lim_(x->2)(x^4-8x)/(x-2)#

Since substituting two in the place of #x# results in #0/0#, we can use the L'Hospital's Rule.

The rule states that: #lim_(x->c)f(x)/g(x)=(f'(c))/(g'(c))# if #f(c)/g(c)# results in an indeterminate form such as #0/0#.

Let's find the derivatives using the power rule: #d/dx[x^n]=nx^(n-1)# where #n# is a constant.

The numerator:

#d/dx(x^4-8x)#

#=>4x^(4-1)-8x^(1-1)#

#=>4x^(3)-8x^(0)#

#=>4x^(3)-8*1#

#=>4x^(3)-8#

The denominator:

#d/dx(x-2)#

#=>1*x^(1-1)-2*0x^(0-1)#

#=>x^(0)-0#

#=>1#

We now have:

#(4x^(3)-8)/1# Substitute 2 in the place of #x#.

#=>(4(2)^(3)-8)#

#=>24#

Therefore, #lim_(x->2)(x^4-8x)/(x-2)=24#

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