Solution verification: $\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$

Question

Find (without L'Hospital):$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$$

My attempt:

$$\lim_{x\to 0}\cosh x=1$$ $$\lim_{x\to 0}e^{2\sin x}\cosh x-1=\lim_{x\to 0}\frac{e^{2\sin x}-1}{2\sin x}\cdot2\sin x=2\lim_{x\to 0}\sin x$$ $$\lim_{x\to 0}(\cosh x-\cos x)=\lim_{x\to 0}(1-\cos x)=\lim_{x\to 0}\frac{1-\cos x}{x^2}\cdot x^2=\frac{1}{2}\lim_{x\to 0}x^2$$ $$\lim_{x\to 0}\sqrt[3]{x(\cosh x-\cos x)}=\lim_{x\to 0}\sqrt[3]{x\cdot\frac{1}{2}x^2}=\frac{1}{\sqrt[3]{2}}\lim_{x\to 0}x$$ $$2\sqrt[3]{2}\lim_{x\to 0}\frac{\sin x}{x}=2\sqrt[3]{2}$$

Is this legitimate?

Answer 1

The key is to use laws of algebra of limits whose primary function is to help evaluate the limit of a complicated function which is composed of many simpler expressions connected by algebraic operations $+, -, \times, /$ given the limits of these simpler expressions. I have described these laws in a simple manner in this answer.

You should have in memory a table of well known standard limits which allow us to deal with the simpler expressions mentioned in last paragraph.

In this question we need the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1=\lim_{x\to 0}\frac{\sin x} {x} $$ and $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}=\lim_{x\to 0}\frac{\cosh x-1}{x^2}$$ First we deal with denominator. We can rewrite it as $$x\sqrt[3]{\frac{\cosh x-1}{x^2}+\frac{1-\cos x} {x^2}}$$ The expression under radical sign tends to $(1/2)+(1/2)=1$ and hence the denominator can be safely replaced by $x$.

Next we deal with numerator. It can be rewritten as $$e^{2\sin x} \cosh x-\cosh x +\cosh x - 1$$ or $$\cosh x (e^{2\sin x} - 1)+\cosh x - 1$$ Thus the desired limit is equal to the limit of $$\cosh x\cdot\frac{e^{2\sin x} - 1}{2\sin x} \cdot 2\cdot\frac{\sin x} {x} +x\cdot\frac{\cosh x - 1}{x^2}$$ which is $$1\cdot 1\cdot 2\cdot 1 +0\cdot\frac{1}{2}=2$$

Answer 2

$$L=\lim_{x \rightarrow 0}\frac{e^{2 \sin x} \cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}= \lim_{x\rightarrow 0} \frac{(1+2x+O(x^2))(1+x^2/2+O(x^4))-1}{{\sqrt[3]{x^3+O(x^5)}}}$$ $$\implies L= \lim_{x\rightarrow 0} \frac{2x+x^2/2+x^3+O(x^4)}{x}=2.$$ Here we have used $\cosh x \approx 1+x^2/2 +O(x^4), \cos x \approx 1-x^2/2+O(x^4), \sin x \approx x|O(x^3),$ $e^{x}\approx 1+x+O(x^2),$ when $x \rightarrow 0$

Answer 3

There is an error in your expansion for the denominator, $$\sqrt[3]{x(\cosh x-\cos x)}\ne \frac{1}{\sqrt[3]{2}}x\>+\>...$$

Instead, it should be $$\sqrt[3]{x(\cosh x-\cos x)}=x +O(x^5)$$

along with the numerator

$$e^{2\sin x}\cosh x-1 = 2x +O(x^2)$$ As a result,

$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}=\lim_{x\to 0}\frac{2x+O(x^2)}{x +O(x^5)}=2$$

Answer 4

As $\cosh(x)=\cos(ix)$ using Prosthaphaeresis Formula

$$\lim_{x\to0}\dfrac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cos h x-\cos x)}} =\lim_{x\to0}\dfrac{e^{2\sin x}(e^x+e^{-x})-2}{2\sqrt[3]{2x\sin\dfrac{x(1+i)}2\sin\dfrac{x(1-i)}2}}$$

$$=\lim_{x\to0} \dfrac{x\left(\dfrac{e^{2\sin x+x}-1}x+\dfrac{e^{2\sin x-x}-1}x\right)}{2x\sqrt[3]{2\cdot\dfrac{\sin\dfrac{x(1+i)}2}{\dfrac{x(1+i)}2}\cdot\dfrac{\sin\dfrac{x(1-i)}2}{\dfrac{x(1-i)}2}}\cdot\dfrac{(1+i)(1-i)}4}$$

$$=\dfrac{2+1+2-1}{2}$$