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I need to find $\lim_{x\to 0}(1+x^{2})^{\cot^2{x}}$.

I started like this:

$\lim_{x\to0}(1+x^{2})^{\cot^2{x}} = \lim_{x\to0}e^{\ln(1+x^{2})^{\cot^2{x}}} = \lim_{x\to0}e^{\cot^2{x} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})}$

Now I'm not sure about this one but can I say:

$\lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})}$?

if so than

$ \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{1 \cdot \ln(1)} = e^0 = 1$

Is this valid? Is there a better way

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    How did you replace $\cos^2 x$ with $x^2$? – 2017-01-14
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    The answer is $e$. See my answer. – 2017-01-14

4 Answers 4

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$$\lim_{x\to 0}(1+x^{2})^{\cot^2{x}}$$

  • If you know that $\frac{\sin x}{x}\to1$ as $x\to0$ (or, equivalently, that $\frac{\tan x}{x}\to1$ as $x\to0$), then you can deduce that $x^2\cot^2x\to1$ as $x\to0$.
    • This first limit can be proved by geometric methods, and is fairly standard.
  • If you know that $\left(1+\frac{1}{n}\right)^n\to e$ as $n\to\infty$, substituting $n=x^{-2}$ allows you to infer that $(1+x^2)^{1/x^2}\to e$ as $x\to0$.

Thus:

$$\lim_{x\to 0}(1+x^{2})^{\cot^2{x}}=\lim_{x\to 0}\left[(1+x^{2})^{1/x^2}\right]^{x^2\cot^2 x}=e^1=e$$

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$$\lim_{x\to 0}(1+x^2)^{\cot^2x}$$ $$=\lim_{x\to 0}(1+x^2)^{\frac {1}{x^2}x^2\cot^2x}$$ $$=\lim_{x\to 0}(1+x^2)^{\frac {1}{x^2}{(\frac{x}{\sin x})}^2\cos^2x}$$ We know $\lim_{x\to 0}\frac{x}{\sin x} = 1$ So $$=\lim_{x\to 0}(1+x^2)^{\frac {1}{x^2}\cos^2x}$$ and $\cos0= 1$. Also $\lim_{x\to 0}(1+x)^\frac{1}{x} = e$.

Thus, $$\implies \lim_{x\to 0}(1+x^2)^{\frac {1}{x^2}}=e$$

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No, you can't say that. $\cos^2 x$ and $x^2$ are not very much alike close to $0$. The limit is actually $e$. But your tag says "limits without L'hospital", and I don't see a way to get $e$ without it.

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Take the log we get $$\cot^2 x \ln(1+x^2)=\cos^2 x \frac{\ln(1+x^2)}{\sin^2 x}$$ write as

$$\cos^2 x \frac{\ln(1+x^2)}{x^2}\frac{x^2}{\sin^2 x}\to 1$$ by standard limits. Easy peasy.

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    How can you tell that $\lim_{x\to 0}\frac{\ln(1+x^2)}{x^2}=1$? Is it: $\lim_{x\to 0}\frac{\ln(1+x^2)}{x^2} = \lim_{x\to 0}\ln((1+x^2)^{\frac{1}{x^2}} )= \lim_{t\to 0}\ln((1+t)^{\frac{1}{t}}) = \ln(e) = 1$? – 2017-01-14
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    @juniven This is just $\frac{ln(1+x)}{x}\to 1$. When you talk about showing this, you must remember that it is one of the basic limits of the theory and depends on the properties unique to $e$. Also when we say $d(\ln x)=\frac{1}{x}$ this is just a restatement of the above limit. So it is really a question of how far back in the limit theory one wants to go. For myself I take it as a basis for all further limit calculations. Its a restatement of $\frac{e^x -1}{x}\to 1$. – 2017-01-14
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    @Rene Schipperus Okay, got the point. – 2017-01-14
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    @juniven Ok, but the comment of S.Peter above is a nice proof. – 2017-01-14
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    @Rene Schipperus Yeah, its nicely done – 2017-01-14
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    The answer is $e$. – 2017-01-14
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    @Gyanshu Yeah but I took the log....so I get one. – 2017-01-14
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    At the end you get $log_e(answer) = 1$. $\therefore$ answer $= e^1$. Remember you took log both sides at the starting of the problem? – 2017-01-14
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    @Gyanshu Oh dear prof, am I going to lose marks ? – 2017-01-14
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    sure if you do that in an exam ;P – 2017-01-14
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    @Gyanshu The fact of the matter is that I shall never take another exam in this life. – 2017-01-14
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    Then you won't loose any marks.Fine? :P – 2017-01-14
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    @Gyanshu Anyway I never cared about marks only my own knowledge. – 2017-01-14