Calculating $\displaystyle{\lim_{x \to 0^+}}{\frac{1}{\sqrt{x}}\Big(e^x + \frac{2\log(\cos(x))}{x^2}}\Big)$

Question

I am struggling to calculate this limit:

$$\displaystyle{\lim_{x \to 0^+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}}$$

I prefer not to use l'Hopital's rule, only when necessary. If possible, solving with the help of these limits:

$\displaystyle{\lim_{x \to 0^+}}{\frac{\log(x + 1)}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{e^x - 1}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{1 - \cos(x)}{x^2}} = \frac{1}{2}$.

Also when I tried solving it using only l'Hopital, it seemed to be very laborious, and also not sure if sufficient to solve it.

Also I prefer not using Taylor theorem and little/big o notation if possible at all.

Here's my (unsuccessful) attempt:

$$\displaystyle{\lim_{x \to 0+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}} = \displaystyle{\lim_{x \to 0^+}}{\frac{e^xx^2 + e^x2x + 2\log{\cos{x}}}{x^2\sqrt{x}}} = \frac{1}{5}\displaystyle{\lim_{x \to 0^+}}{\frac{e^x(x^2 + 2x) - 2tg{x}}{x\sqrt{x}}} = ...$$

Problem is, each time I use l'Hopital, it doesn't seem to simplify limit in any way.

Thanks.

Answer 1

The known limits in the question are not enough to find the limit in the question. Some application of l'Hopital or Taylor's theorem will be needed. This answer shows how to compute the limit using l'Hopital.

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The trick here to apply l'Hopital more easily is to compute two limits separately instead of one.

$$\begin{aligned}\lim_{x \to 0^+}\frac{e^x -1}{x}&\color{blue}{=\lim_{x \to 0^+}\frac{e^x}{1}}=1\\ \ \\ \lim_{x \to 0^+}{\frac{1 + \frac{2\log(\cos(x))}{x^2}}{x}} &=\lim_{x \to 0^+}\frac{x^2+2\log(\cos(x))}{x^3}\\ &\color{blue}{=\lim_{x \to 0^+}\frac{2x+2\frac{-\sin(x)}{\cos(x)}}{3x^2}}\\ &\color{blue}{=\lim_{x \to 0^+}\frac{2+2\frac{-1}{\cos^2(x)}}{6x}}\\ &=\lim_{x \to 0^+}\frac{-2\sin^2(x)}{3x\cos^2(x)}\\ &=\lim_{x \to 0^+}\frac{-2\sin^2(x)}{3x}\\ &\color{blue}{=\lim_{x \to 0^+}\frac{-4\cos(x)\sin(x)}{3}}\\ &=0 \end{aligned}$$ Adding both limits, we see that $\lim_{x \to 0^+}{\frac{e^x+ \frac{2\log(\cos(x))}{x^2}}{x}}=1$. Since $\lim_{x \to 0^+}\frac{x}{\sqrt x}=0$, the limit in the title is $0$.

_{Expressions in blue are obtained by l'Hopital.}

_{Applying l'Hopital repeatedly is akin to using Taylor's theorem. However, Taylor' theorem is much clearer and handier. So, here is a piece of unsolicited advice. Get used to Taylor's theorem and apply it more!}

Answer 2

Note from Taylor's theorem that

$$\begin{align} \frac{2\log(\cos(x))}{x^2}&=\frac{2\log(1-2\sin^2(x/2))}{x^2}\\\\ &=-1+O(x^2) \end{align}$$

Then, we see that

$$\frac{e^x+\frac{2\log(\cos(x))}{x^2}}{\sqrt x}=\sqrt{x}+O(x^{3/2})$$

from which we conclude the limit is zero.

Answer 3

As the OP asks, an attempt to do it, using only \begin{align} \log(1+u)&=u+o(u),\\ e^x &= 1 + x + o(x),\\ \cos(x) &= 1 -\frac{x^2}{2} + o(x^2) . \end{align}

Now, as $x \to 0$, the best we can deduce is:
\begin{align} \cos x &= 1 - \frac{x^2}{2} + o(x^2) \\ \log \cos x &= \log(1-(1-\cos x)) = -(1-\cos x) + o\big(1-\cos x\big) \\ &= -1+\cos x + o\big(x^2\big) = -1 + \big(1-\frac{x^2}{2} + o(x^2)\big)+ o\big(x^2\big) \\ &=-\frac{x^2}{2}+o(x^2) \\ \frac{2\log \cos x }{x^2} &= -1+o(1) \\ e^x &= 1+x+o(x) \\ e^x+\frac{2\log \cos x }{x^2} &= 1+x+o(x)-1+o(1) =o(1) \\ \frac{e^x+\frac{2\log \cos x }{x^2}}{\sqrt{x}} &= o(x^{-1/2}) \end{align} This is not enough to get the answer.

From this we can see how to get a counterexample. Instead of $\cos(x)$, use $c(x) := 1-\frac{x^2}{2}+x^{5/2}$. This still satisfies $c(x) = 1-\frac{x^2}{2}+o(x^2)$, but

$$ \frac{e^x+\frac{2\log c(x) }{x^2}}{\sqrt{x}} \to 2 $$ and not $0$.

Answer 4

$\lim_{x \to 0^{+}}\dfrac{1}{\sqrt x}\bigg(e^{x}+\dfrac{2\log(1+\cos x-1)}{(\cos x-1)}\cdot\dfrac{(\cos x-1)}{x^2}\bigg)$

$\implies \lim_{x \to 0^{+}}\dfrac{1}{\sqrt x}\bigg(e^{x}-1\bigg)$

$\implies \lim_{x \to 0^{+}}\dfrac{1}{\sqrt x}\bigg(\dfrac{e^{x}-1}{x}\bigg)\cdot x\implies\lim_{x\to 0^{+}}{\sqrt x}=0$

Note:

$(1)\;\lim_{x\to 0^{+}} \dfrac{\log(1+\cos x-1)}{(\cos x-1)}=1$

$(2)\; \lim_{x\to 0^{+}}\dfrac{(\cos x-1)}{x^2}=\dfrac{-1}{2}$

$(3)\;\lim_{x\to 0^{+}}\dfrac{e^{x}-1}{x}=1$

Answer 5

We have that

$$\frac{1}{\sqrt{x}} \left(e^x + \frac{2\log(\cos(x))}{x^2}\right)=\frac{1}{\sqrt{x}} \left(e^x -1+1+ \frac{2\log(\cos(x))}{x^2}\right)=$$

$$= \frac{e^x -1}{\sqrt{x}}+\frac{1}{\sqrt{x}} \left(1+ \frac{2\log(\cos(x))}{x^2}\right)\to 0$$

indeed

$$\frac{e^x -1}{\sqrt{x}} =\sqrt{x}\cdot\frac{e^x -1}{x} \to 0$$

and

$$\frac{1}{\sqrt{x}} \left(1+ \frac{2\log(\cos(x))}{x^2}\right)=\sqrt{x} \left( \frac{x^2+\log(\cos^2(x))}{x^3}\right)\to 0$$

since using the results given here

$$\frac{x^2+\log(\cos^2(x))}{x^3}=\frac{x^2-\sin^2x+\sin^2x+\log(1-\sin^2(x))}{x^3}=$$

$$=(x+\sin x)\frac{x-\sin x}{x^3}+x\frac{\sin^2x+\log(1-\sin^2(x))}{\sin^4 x}\frac{\sin^4 x}{x^4}\to 0\cdot\frac 16+0\cdot \left(-\frac12\right)\cdot 1 =0$$