2
$\begingroup$

I correctly solved to following limit like this:

$$\lim_{x\to\infty} \left(\frac{1}{x^2}\right)^{\frac{2x}{x+1}}$$ $$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2x}{x(1+\frac{1}{x})}}$$ $$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2}{(1+\frac{1}{x})}}$$ $$ = 0^2 = 0$$

However, I usually use the following formula for these kind of limits: $\lim_{x\to\infty} (1+\frac{1}{v})^v = e$ This is simple one that doesn't really need this extra complexity, but anyway. The problem is that I get a different and wrong result:

$$\lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2x}{x+1}}$$ $$= \lim_{x\to\infty} (1 + \frac{1}{x^2} - 1)^{\frac{2x}{x+1}}$$ $$= \lim_{x\to\infty} (1 + \frac{1-x^2}{x^2})^{{\frac{2x}{x+1} \frac{1-x^2}{x^2} \frac{x^2}{1-x^2}}}$$ $$= e^{ \lim_{x\to\infty} {{\frac{2x}{x+1} \frac{1-x^2}{x^2}}}}$$ $$= e^{ \lim_{x\to\infty} {{\frac{2x-2x^3}{x^3+x^2}}}} = e^{ \lim_{x\to\infty} {{\frac{-2x^3}{x^3}}}} = e^{-2}$$

$\endgroup$
1
  • $\begingroup$ That's impressive latex. Just one suggestion: you might want to use \left and \right to format a bit prettier, I edited the first limit that way, so you can see what I'm talking about. $\endgroup$ Jul 26 '16 at 19:04
7
$\begingroup$

$$\frac{1-x^2}{x^2}$$ does not approach zero as $x$ approaches infinity, so you cannot simply write that $$\lim_{x\rightarrow\infty}\left(1+\frac{1-x^2}{x^2}\right)^{\frac{x^2}{1-x^2}}=e$$

$\endgroup$
1
  • $\begingroup$ Oh ok, didn't know that requirement. My book just listed the formula without further information. $\endgroup$
    – Aaron
    Jul 26 '16 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.