Do you have rules for differentiation or are you using the definition? (a definition)

**Important:** You'll get the same answer either way, but I don't know where you are in your study of calculus.

If you are using a definition, it's probably one of the two below:

.

#f(x)=4x^3+2x-3#

**Solution 1:**

Using definition: #lim_(xrarr2)(f(x)-f(2))/(x-2)#

#lim_(xrarr2)(f(x)-f(2))/(x-2)=lim_(xrarr2)([4x^3+2x-3]-[4(2)^3+2(2)-3])/(x-2)#

#=lim_(xrarr2)(4x^3+2x-3-33)/(x-2)=lim_(xrarr2)(4x^3+2x-36)/(x-2)#

Trying to evaluate this limit by substitution gives indeterminate form: #0/0#. But don't give up hope!

Because #2# is a zero of the polynomial numerator, we can be sure that #(x-2)# is a factor.

#4x^3+2x-36=(x-2)(4x^2+8x+18)# (by division or by trial and error or, perhaps, by grouping)

Resuming:

#lim_(xrarr2)(f(x)-f(2))/(x-2)=lim_(xrarr2)(4x^3+2x-36)/(x-2)#

#=lim_(xrarr2)((x-2)(4x^2+8x+18))/(x-2)=lim_(xrarr2)(4x^2+8x+18)#

#=4(2)^2+8(2)+18=16+16+18=50#

**Solution 2:**

Use the definition: #lim_(hrarr0)(f(2+h)-f(2))/h#

#lim_(hrarr0)(f(2+h)-f(2))/h#

#=lim_(hrarr0)([4(2+h)^3+2(2+h)-3]-[4(2)^3+2(2)-3])/h#

#=lim_(hrarr0)([4(8+12h+6h^2+h^3)+2(2+h)-3]-[33])/h#

#=lim_(hrarr0)([32+48h+24h^2+4h^3+4+2h-3]-[33])/h#

#=lim_(hrarr0)(50h+24h^2+4h^3)/h#

#=lim_(hrarr0)(50+24h+4h^2)=50#