Find $\lim_{x\to0}(1+x^{2})^{\cot^2{x}}$ without lhopital

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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

• 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:

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

Answer 4

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