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How to calculate following limit without using L'Hospital rule?

$$\lim_{x \rightarrow\infty}\left(\frac{(2+x)^{40}(4+x)^{5}}{(2-x)^{45}} \right)$$

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3 Answers 3

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$\displaystyle \lim_{x \rightarrow\infty}\left(\frac{(2+x)^{40}(4+x)^{5}}{(2-x)^{45}} \right)=\lim_{x \rightarrow\infty}\left(\frac{(2/x+1)^{40}(4/x+1)^{5}}{(2/x-1)^{45}} \right)=-1$

(In the first step we divide the numerator and denominator by $x^{45}$)

As we know $\lim_{x\to \infty }\frac{c}{x}=0,c$ is constant .So $\lim_{x\to \infty}(\frac{c}{x}+d)^k=d^k$(for ($k\in N$)and $c,d\in R$).

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  • $\begingroup$ nice step still its difficult to solve $\endgroup$ Mar 20, 2013 at 15:21
  • $\begingroup$ @Abhra, specify the nature of $c$ to make things more perfect $\endgroup$ Mar 20, 2013 at 15:25
  • $\begingroup$ Ok got it thank you very much $\endgroup$ Mar 20, 2013 at 15:25
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Putting $h=\frac1x$

As $x\to\infty ,h\to0$

$$\lim_{x \rightarrow\infty}\left(\frac{(2+x)^{40}(4+x)^{5}}{(2-x)^{45}} \right)$$

$$=\lim_{h \rightarrow 0}\left(\frac{(2+\frac1h)^{40}(4+\frac1h)^{5}}{(2-\frac1h)^{45}} \right)$$

$$=\lim_{h \rightarrow 0}\left(\frac{(2h+1)^{40}(4h+1)^{5}}{(2h-1)^{45}} \right)$$ multiplying the numerator & the denominator by $h^{45}$ as $h\ne0$ as $h\to0$

$$=\frac{1\cdot1}{(-1)^{45}}$$

$$=-1$$

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Just note that

$$ \left(\frac{(2+x)^{40}(4+x)^{5}}{(2-x)^{45}} \right) \sim \left(\frac{(x)^{40}(x)^{5}}{(-x)^{45}} \right) .$$

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