# Calculate `lim_(x->oo) ((2x+1)^5*(3x+1)^5)/(2x+6)^10 `

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You may substitute `oo ` for x in equation under limit such that:

`lim_(x->oo) (2x+1)^5*(3x+1)^5/(2x+6)^10 = oo/oo`

You may use binomial theorem such that:

`(a+b)^n = C_n^0 a^n + C_n^1 a^(n-1)b +.. +C_n^k a^(n-k)b^k ` `+ ... + C_n^n b^n`

Reasoning by analogy yields:

`(2x+1)^5 =C_5^0 (2x)^5 + C_5^1 (2x)^4 + C_5^2 (2x)^3 + C_5^3 (2x)^2 + C_5^4 (2x) + C_5^5`

Substituting 1 for `C_5^0` and `C_5^5` , 5 for `C_5^1 ` and `C_5^4` , 10 for `C_5^2` and `C_5^3` yields:

`(2x+1)^5 = 32x^5 + 5*16 x^4 + 10*8 x^3 + 10*4 x^2 + 5*2 x + 1`

`(2x+1)^5 = 32x^5 + 80x^4 +80 x^3 +40 x^2 +10 x + 1`

` (2x+1)^5 = C_5^0 (3x)^5 + C_5^1 (3x)^4 + C_5^2 (3x)^3 + C_5^3 (3x)^2 + C_5^4 (3x) + C_5^5`

`(2x+1)^5 = 243x^5 + 405x^4 + 270x^3 + 90x^2 + 15x + 1`

`(2x+6)^10 = C_10^0 (2x)^10 + C_10^1 (2x)^9 +...+ C_10^10 6^10`

Hence, using the binomial expansions yields:

`lim_(x->oo) ((32x^5 + 80x^4...)(243x^5 + 405x^4 + ...))/(32x^10 + ...)`

You should force factors `32x^5, 243x^5` and `1024x^10` such that:

`lim_(x->oo) (32*243x^10(1 + 80/x + ...)(1 + 405/x + ...))/(1024x^10(1 + ...))`

Reducing by `32x^10` yields:

`lim_(x->oo) 243(1 + 80/x + ...)(1 + 405/x + ...)/(32(1 + ...))`

Since `lim_(x->oo) 80/x = lim_(x->oo) 405/x = ... = 0` , hence, evaluating the limit yields:

`lim_(x->oo) (2x+1)^5*(3x+1)^5/(2x+6)^10 = 243/32`

**Hence, evaluating the given limit forcing the indicated factors yields `lim_(x->oo) (2x+1)^5*(3x+1)^5/(2x+6)^10 = 243/32.` **