Limit problem

limit problem x tends to 3 ((3+2x)^5-243)/(3x)

 

Perhaps I misunderstand what you are asking???

 

\(\displaystyle \lim_{x\rightarrow3}\;\;\frac{ (3+2x)^5-243}{3x}\\ =\frac{ (3+2*3)^5-243}{3*3}\\ =\frac{ 9^5-243}{9}\\ =\frac{ 9^5-9*27}{9}\\ =\frac{ 9^4-27}{1}\\ =6534\)

Find the following limit:
lim_(x->3) ((2 x+3)^5-243)/(3 x)

 

((2 x+3)^5-243)/(3 x) = 2/3 (16 x^4+120 x^3+360 x^2+540 x+405):
lim_(x->3) 2/3 (16 x^4+120 x^3+360 x^2+540 x+405)

 

lim_(x->3) 2/3 (16 x^4+120 x^3+360 x^2+540 x+405) = 2/3 (405+540 3+360 3^2+120 3^3+16 3^4) = 6534:
Answer: |6534

I suspect this should be find the limit as x tends to zero.

 

If so:

 

(3 + 2x)^5 → 3^5 + 5*3^4*2x + O(x^2) → 243 + 10*3^4*x + O(x^2)

 

So:

 

((3+ 2x)^5 - 243)/(3x) → 10*3^3 + O(x) → 270 + O(x)

 

In the limit as x tends to zero, this tends to 270

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