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Let $0<x<1$ be a real number and let $a$ be a positive integer.

I want to compute $$\lim_{t\to 0} \frac{ (x+\frac{t}{x^2})^a}{t^a}.$$

The only thing I can come up with is to use L'Hopital's rule.

Differentiating the denominator and numerator $a$ times I end up with

$$\lim_{t\to 0} \frac{ (x+\frac{t}{x^2})^a}{t^a} = \frac{1}{x^{2a}}.$$

Is this correct?

Thnx for your help!

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1  
Is t supposed to be equal to r? If not, you have a constant on the numerator. EDIT: Misread the question. There's a t on the numerator. –  josh Oct 6 '11 at 19:11
3  
HINT: For integer $a$ or positive $t$, $(x+t/x^2)^a/t^a = ((x+t/x^2)/t)^a = (x/t+1/x^2)^a$. –  Sasha Oct 6 '11 at 19:16
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Be careful. Can you use L'Hopital? Do you really get "0/0"? Also, even if L'Hopital applies, you must differentiate with respect to "t" not "x" ("t" is the limiting variable). –  Bill Cook Oct 6 '11 at 19:20

1 Answer 1

up vote 4 down vote accepted

Since $x$ is fixed and $t\to0$, $x+t/x^2\to x$ hence $(x+t/x^2)^a$ converges to $x^a$. The ratio $\dfrac1{t^a}$ converges to $\pm\infty$, depending on whether $t\to0^+$ or $t\to0^-$, and on the parity of the positive integer $a$, hence the whole thing converges to $\pm\infty$ as well.

If $a$ is even, the limit is $+\infty$. If $a$ is odd, the limit does not exist, but the limit from the left (that is, when $t\to0^-$) is $-\infty$ and the limit from the right (that is, when $t\to0^+$) is $+\infty$.

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I want to learn the difference between say the limit is $+\infty$ and saying that the limit does not exist. Thanks. –  Emmad Kareem Oct 6 '11 at 21:51

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