Solve $\lim_{x\to 0} (\ln (x+e))^{\cot x}$ without l'Hôpital

Question

$$\lim_{x\to 0} (\ln (x+e))^{\cot x}$$

Since we have an indeterminate form of $1^{\infty }$, we should simplify in a way that we can transform the expression into e to the power of something, but I can't find a way.

The use of l'Hôpital's rule or series is prohibited.

Answer 1

First of all, you want to compute the limit of the logarithm of your function: $$ \lim_{x\to0}\cot x\ln(\ln(x+e)) $$ The first step is almost obvious: note that $$ \lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1 $$ so you can reduce to computing $$ \lim_{x\to0}\frac{\ln(\ln(x+e))}{x} $$ because if this limit exists it will be equal to the one you want to compute.

How do we do this one? Let's do the substitution $x=et$, so it becomes $$ \lim_{t\to0}\frac{\ln(\ln e+\ln(1+t))}{et}= \lim_{t\to0}\frac{\ln(1+\ln(1+t))}{et} $$ Not yet in the best form. Let's do a new substitution: $\ln(1+t)=u$, that is, $t=e^u-1$, so we get $$ \frac{1}{e}\lim_{u\to0}\frac{\ln(1+u)}{e^u-1}= \frac{1}{e}\lim_{u\to0}\frac{\ln(1+u)}{u}\frac{u}{e^u-1}=… $$

Since $$\lim_{u\to0}\frac{\ln(1+u)}{u}=1\qquad\lim_{u\to0}\frac{u}{e^u-1}=1$$ are both known limits, you can conclude that $$\lim_{x\to0}\cot x\ln(\ln(x+e))=\frac{1}{e}$$ Therefore $$\lim_{x\to0}(\ln(x+e))^{\cot x}=e^{1/e}$$

Of course, using l'Hôpital is easier and doesn't require imagination: $$ \lim_{x\to0}\frac{\ln(\ln(x+e))}{\tan x}= \lim_{x\to0}\frac{\dfrac{1}{\ln(x+e)}\dfrac{1}{x+e}}{1+\tan^2x}=\frac{1}{e} $$

Answer 2

Rewrite it as $$(\ln(x+e))^{\cot(x)}=\exp\left[x\cot(x)\frac{\ln(\ln(x+e))}x\right]$$

And use the known limit

$$\lim_{x\to0}\frac{\sin(x)}x=1$$

To get

$$\lim_{x\to0}x\cot(x)=\lim_{x\to0}\cos(x)\left(\frac{\sin(x)}x\right)^{-1}=1\cdot1^{-1}=1$$

and then use the definition of the derivative to compute

$$\lim_{x\to0}\frac{\ln(\ln(x+e))}x=\frac d{dx}\ln(\ln(x+e))\bigg|_{x=0}$$

Evaluating the derivative with chain rule and

$$\frac d{dx}\ln(x)=\frac1x$$

which should give you a final result of

$$\lim_{x\to0}(\ln(x+e))^{\cot(x)}=e^{1/e}$$

Answer 3

You have \begin{align} \ln(e+x)^{\cot x}&=\ln\left(e(1+\frac xe)\right)^{\frac{\cos x}{\sin x}}=\left(1+\ln(1+\frac xe)\right)^{\frac{\cos x}{\sin x}}\\ \ \\ &=\left(1+\frac xe+o(x^2) \right)^{\frac{\cos x}{\sin x}}\\ \ \\ &=\exp\left(\frac{\cos x}{\sin x}\,\ln\left(1+\frac xe+o(x^2) \right) \right)\\ \ \\ &=\exp\left(\frac{\cos x}{\sin x}\,\left(\frac xe+o(x^2)\right)\right)\\ \ \\ &=\exp\left(\frac{1+o(x^2)}{x+o(x^3)}\,\left(\frac xe+o(x^2)\right)\right)\\ \ \\ &=\exp\left(\frac{1+o(x^2)}{1+o(x^2)}\,\left(\frac 1e+o(x)\right)\right)\\ \ \\ &\xrightarrow[x\to0]{}\exp\left(\frac1e\right)=e^{1/e}. \end{align} The relevant Taylor polynomials used above are $$ \ln(1+x)=x+o(x^2),\ \ \cos x=1+o(x^2),\ \ \ \sin x=x+o(x^3), $$ together with $a^b=e^{b\ln a}$ (which is the definition of $a^b$)

Answer 4

Note that:

$$\ln(x+e)=\ln e +\ln\left(1+\frac{x}{e}\right)=1+\frac{x}{e}+o(x)$$

Thus:

$$(\ln (x+e))^{\cot x}=\left[\left(1+\frac{x}{e}+o(x)\right)^\frac1x\right]^\frac{x}{tanx}\to e^{\frac1e}$$

Answer 5

Taking the natural log you have $$\cot x \ln \ln(x+e))=$$ $$\cos x \frac{x}{\sin x} \frac{\ln \ln (x+e)}{\ln (x+e)-1}\frac{\ln(x+e)-1}{x}$$ The initial factors all limit to $1$, as for the last, $$\frac{\ln(x+e)-1}{x}= \frac{\ln(x+e)-\ln e}{x}=\frac{\ln(\frac{x}{e}+1)}{x}$$ $$=\frac{1}{e}\frac{\ln(\frac{x}{e}+1)}{x/e}\to\frac{1}{e}$$