Question

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

• A. $-1$
• B. $1$
• C. $16$
• D. $32$

Answer 1

The correct option is A $-1$

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

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

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

when $x\to \infty ,\frac{2}{x}\to 0$

$=\frac{\left(1{)}^{40}\right(1{)}^5}{(-1{)}^{45}}$

$=\frac{(1{)}^{45}}{(-1{)}^{45}}$
$=-1$