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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Limits and Derivatives

$\lim\limits_{x\to \infty}\big(\large\frac{x+6}{x+1}\big)^{x+4}$=

$(a)\;e\qquad(b)\;e^2\qquad(c)\;e^5\qquad(d)\;e^4$

1 Answer

$\lim\limits_{x\to \infty}\big(\large\frac{x+6}{x+1}\big)^{x+4}$$=\lim\limits_{x\to \infty}\bigg(\big[1+\large\frac{5}{x+1}\big]^{\Large\frac{x+1}{5}}\bigg)^{5\large\frac{(x+4)}{x+1}}$
Using $\lim\limits_{x\to \infty}\big(1+\large\frac{1}{x}\big)^x$$=e$
$\lim\limits_{e^{x\to\infty}}5\big(\large\frac{x+4}{x+1}\big)=$$e^5\lim\limits_{x\to\infty}\big(\large\frac{1+4/x}{1+1/x}\big)$
$\Rightarrow e^5$
Hence (c) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 
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