Another combined limit

Question

I've tried to get rid of those logarithms, but still, no result has came. $$\lim_{x\to 0 x \gt 0} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}$$ Please help

Answer 1

Using asymptotics for $\ x\to0$ we have: $$\ \sqrt{1+x^2}≈1+\frac{x^2}{2}$$ $$\ \ln(1+x)≈x$$ $$\ \cos(x)≈1-\frac{x^2}{2}$$ So your limit gets: $$\ \lim_{x\to0^+}\frac{\ln(x+\sqrt{1+x^2})}{\ln(\cos x)}≈\lim_{x\to0^+}\frac{\ln(1+x+\frac{x^2}{2})}{\ln(1-\frac{x^2}{2})}≈$$ $$≈\lim_{x\to0^+}\frac{x+\frac{x^2}{2}}{-\frac{x^2}{2}}=-\infty$$

Answer 2

HINT:

To get rid of the $ln$ in the denominator, remember that $\ln \cos x= \frac{1}{2} \ln(1-\sin^2(x))$.

As for the $ln$ in the nominator, you'll have to calculate the limit: $\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}$ wich is easier (it's zero).

SOLUTION:

\begin{align} \lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}} &=2\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln(1-\sin^2(x))}\tag{1}\\ &=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \frac{x^2}{\sin^2(x)} \cdot \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{2}\\ &=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \lim_{x\to 0^+} \frac{x^2}{\sin^2(x)} \cdot \lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{3}\\ &=2\cdot (-1) \cdot 1 \cdot \infty \tag{4}\\ &=-\infty \tag{5}\\ \end{align}

Explanation:

$(1)$: $\frac{1}{2} \ln(1-\sin^2(x))=\frac{1}{2} \ln(\cos^2(x))=\frac{1}{2} \cdot 2\ln(\cos x)=\ln \cos x$

$(4)$: $\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} =\lim_{x\to 0^+} \frac{1}{\ln(1-\sin^2(x))^{1/\sin^2(x)}}=\lim_{y\to 0^+} \frac{1}{\ln(1-y)^{1/y}}=\frac{1}{\ln(\frac{1}{e})}=-1 $

and

$\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}=\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}=\ln(\infty)=\infty$

Answer 3

Outline: Our expression, for $x\ne 0$, is equal to $$\frac{\ln(x+\sqrt{x^2+1})}{x} \cdot\frac{x}{\ln(\cos x)}.$$

Note that $$\lim_{x\to 0} \frac{\ln(x+\sqrt{x^2+1})-0}{x}$$ is by definition the derivative of $\ln(x+\sqrt{x^2+1})$ at $x=0$. It is easy to compute that derivative. The only important thing is that it is positive.

Now we examine what happens to $\frac{x}{\ln(cos x)}$ as $x\to 0$. It is easier to look at what happens to $\frac{\ln(cosx)}{x}$. Again, think derivative.

Answer 4

Using $\lim\limits_{x\to0}\frac{\log(1+x)}x=1$ and $\lim\limits_{x\to0}\frac{\sin(x)}x=1$,

$$ \begin{align} &\lim_{x\to0^+}\frac{\log\left(x+\sqrt{x^2+1}\right)}{\log(\cos(x))}\\ &=\lim_{x\to0^+}\frac{\log\left(1+x+\sqrt{x^2+1}-1\right)}{\log(1+\cos(x)-1)}\\ &=\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\ &=\lim_{x\to0^+}\frac{x+\frac{x^2}{\sqrt{x^2+1}+1}}{\frac{\sin^2(x)}{\cos(x)+1}} \cdot\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{x+\frac{x^2}{\sqrt{x^2+1}+1}} \cdot\lim_{x\to0^+}\frac{\frac{\sin^2(x)}{\cos(x)+1}}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\ &=\lim_{x\to0^+}\frac1x \cdot\lim_{x\to0^+}\frac{x^2}{\sin^2(x)} \cdot\lim_{x\to0^+}\frac{1+\frac{x}{\sqrt{x^2+1}+1}}{\frac1{\cos(x)+1}}\cdot1\cdot-1\\[6pt] &=\lim_{x\to0^+}\frac1x\cdot1\cdot2\cdot1\cdot-1\\[18pt] &=-\infty \end{align} $$

Answer 5

It can be computed this way $$ \frac{\ln (x+\sqrt{x^{2}+1})}{\ln {(\cos {x})}}=\frac{\ln (1+(x-1)+\sqrt{% x^{2}+1})}{(x-1)+\sqrt{x^{2}+1}}\cdot \frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}% \cdot \frac{x^{2}}{(\cos x)-1}\cdot \frac{(\cos x)-1}{\ln (1+\cos x-1)} $$

$$ \lim_{x\rightarrow 0^{+}}\frac{\ln (1+(x-1)+\sqrt{x^{2}+1})}{(x-1)+\sqrt{% x^{2}+1}}=\lim_{y\rightarrow 0}\frac{\ln (1+y)}{y}=1,\ \ \ \ \ \ y=(x-1)+% \sqrt{x^{2}+1} $$ $$ \lim_{x\rightarrow 0^{+}}\frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}\overset{H-Rule}{=% }\lim_{x\rightarrow 0^{+}}\frac{1+\frac{x}{\sqrt{x^{2}+1}}}{2x}=\frac{1}{% 0^{+}}=+\infty $$ $$ \lim_{x\rightarrow 0^{+}}\frac{x^{2}}{(\cos x)-1}=-2. $$ $$ \lim_{x\rightarrow 0^{+}}\frac{(\cos x)-1}{\ln (1+\cos x-1)}% =\lim_{z\rightarrow 0}\frac{z}{\ln (1+z)}=1,\ \ \ \ \ \ z=(\cos x)-1 $$ Therefore $$ \lim_{x\rightarrow 0^{+}}\frac{\ln (x+\sqrt{x^{2}+1})}{\ln {(\cos {x})}}% =1\cdot (+\infty )\cdot (-2)\cdot 1=-\infty $$

EDIT. It is possible to compute the second limit without using L'Hospital's rule. Indeed, $$ \begin{eqnarray*} \lim_{x\rightarrow 0^{+}}\frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}} &=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)+\sqrt{x^{2}+1}\right) \left( (x-1)-\sqrt{x^{2}+1}\right) }{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\ &=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)^{2}-(x^{2}+1)\right) }{% x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) }=\lim_{x\rightarrow 0^{+}}\frac{-2x% }{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\ &=&\lim_{x\rightarrow 0^{+}}\frac{-2}{x\left( (x-1)-\sqrt{x^{2}+1}\right) }=% \frac{-2}{0^{+}\cdot (-2)}=\frac{1}{0^{+}}=+\infty . \end{eqnarray*} $$

Answer 6

Hint: Let $x=\sinh t$, as $t\to0^+.$ Then, since $1+\sinh^2t=\cosh^2t,$ and $\cosh t\pm\sinh t=e^{\large\pm t},~$ the numerator simply becomes t. As for the denominator, use $\cos x\simeq1-\dfrac{x^2}2~,$ in conjunction with $\ln(1+u)\simeq u$.