Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$

Question

Solve without L'Hopital's rule:

$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$

My work:

$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$

$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$

All of my attempts failed.

$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$

Source: Matematička analiza 1, 2. kolokvij

Answer 1

$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}={1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\cdot{x\over\tan(2x)}\cdot{\cosh(3x^2)e^{8x^3}-1\over x^3}$$

and

$${\cosh(3x^2)e^{8x^3}-1\over x^3}={e^{8x^3}-1\over x^3}+e^{8x^3}{\cosh(3x^2)-1\over x^3}={e^{8x^3}-1\over x^3}+{e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x$$

Now it's easy to see that

$${1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\to{1\over1+1}={1\over2}$$

$${x\over\tan(2x)}={\cos x\over2}\cdot{2x\over\sin(2x)}\to{1\over2}\cdot1={1\over2}$$

$${e^{8x^3}-1\over x^3}=8\cdot{e^{8x^3}-1\over8x^3}\to8\cdot1=8$$

and

$${e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x={e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over3x^2\right)^2(9x)\to{1\over1+1}\cdot1^2\cdot0=0$$

and thus

$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}\to{1\over2}\cdot{1\over2}(8+0)=2$$

Answer 2

Using Taylor series from the strating point.

Use the standard series of $\cosh(t)$ and make $t=3x^2$ gives $$\cosh(x^3)=1+\frac{9 x^4}{2}+O\left(x^{8}\right)$$ Using the binomial expansion or Taylor series again $$\sqrt{\cosh(x^3) }=1+\frac{9 x^4}{4}+O\left(x^{8}\right)$$ Use the standard series of $e^t$ and make $t=4x^3$ gives $$e^{4x^3}=1+4 x^3+O\left(x^6\right)$$ Use the standard series of $\tan(t)$ and make $t=2x$ gives $$\tan(2x)=2 x+\frac{8 x^3}{3}+O\left(x^5\right)$$ Combining everything $$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=\frac {4 x^3+\frac{9 x^4}{4}+O\left(x^6\right) } {2 x^3+\frac{8 x^5}{3}+O\left(x^7\right) }$$ Now, long division to get $$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=2+\frac{9 x}{8}-\frac{8 x^2}{3}+O\left(x^3\right)$$

Use your pocket calculator with $x=0.1$; you should get $2.08865$ while the above truncated expansion gives $2.08583$.

Answer 3

\begin{align*} \dfrac{\sqrt{\cosh(3x^{2})}e^{4x^{3}}-1}{x^{2}\tan(2x)}&=\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}\dfrac{2x}{\tan(2x)}\dfrac{1}{\sqrt{\cosh(3x^{2})}e^{4x^{3}}+1}, \end{align*} so the issue is to compute \begin{align*} \lim_{x\rightarrow 0}\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}. \end{align*} Note that \begin{align*} \dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}=\dfrac{(e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2}{2(2x^{3})}. \end{align*} If you are allowed to use power series, then \begin{align*} (e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2\approx 16x^{3}, \end{align*} so the cancellation with the $x^{3}$ in the denominator is safe.

Answer 4

You need to make use of the following standard limits $$\lim_{t\to 0}\frac{e^t-1}{t}=1,\lim_{t\to 0}\frac{\tan t} {t} =1,\lim_{t\to 0}\frac{\sinh t} {t} =1,\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}$$ Using the second limit above the limit in question is equal to the limit of $$\frac{\sqrt{\cosh(3x^2)}e^{4x^3}-1}{2x^3}$$ Adding and subtracting $e^{4x^3}$ in numerator we can split the above fraction as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{2x^3}+\frac{e^{4x^3}-1}{2x^3}$$ and this can be further rewritten as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{\cosh(3x^2)-1}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2\cdot\frac{e^{4x^3}-1}{4x^3}$$ The limit of above expression is equal to that of $$1\cdot\frac{1}{2}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2$$ via the standard limits mentioned at the start of the answer. Next we can use the identity $$\cosh t =1+2\sinh^2(t/2)$$ to rewrite the first fraction as $$\frac{\sinh^2(3x^2/2)}{(3x^2/2)^2}\cdot \frac{(3x^2/2)^2}{2x^3}$$ which tends to $1^2\cdot 0=0$. The desired limit is thus $2$.

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A combination of algebraic manipulation and standard limits is sufficient to solve most limit problems encountered in a typical calculus course.