Question

$\underset{x\to \infty }{lim}\frac{(2x+1{)}^{40}(4x-1{)}^5}{(2x+3{)}^{45}}$

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Answer 1

We have,

$\underset{x\to \infty }{lim}\frac{(2x+1{)}^{40}(4x-1{)}^5}{(2x+3{)}^{45}}$

Degree of polynomial in numerator and denominator is same and it is 45

Divide of numerator and denomiantor by $x^{45}$

$\underset{x\to \infty }{lim}\frac{\frac{(2x+1{)}^{40}(4x-1{)}^5}{x^{45}}}{\frac{(2x+3{)}^{45}}{x^{45}}}$

$\underset{x\to \infty }{lim}\frac{{\left(\frac{2x+1}{x}\right)}^{40}{\left(\frac{4x-1}{x}\right)}^5}{{\left(\frac{2x+3}{x}\right)}^{45}}$

$\underset{x\to \infty }{lim}\frac{{(2+\frac{1}{x})}^{40}{(4-\frac{1}{x})}^5}{{(2+\frac{3}{x})}^{45}}$

=$\frac{2^{40}\times 4^5}{2^{45}}$

=$\frac{2^{40}\times 2^{10}}{2^{45}}=2^5=32$