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:

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

  • $\begingroup$ 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

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

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

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