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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?

PinkyWay
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  • You need to evaluate limit by using theorems meant for evaluating limits and nothing else. Just as $\sqrt{a+b}\neq \sqrt{a} +\sqrt{b} $ but $a(b+c) = ab+ac$, there are things which are and aren't allowed in calculus. – Paramanand Singh Jan 15 '20 at 05:46
  • Deal with numerator by adding subtracting $\cosh x$ (or $e^{2\sin x} $ if you prefer that). For denominator you need to add subtract $1$ to $\cosh x - \cos x$. – Paramanand Singh Jan 15 '20 at 05:49
  • @ParamanandSingh, we go superficially through so much in our exercises. There are many things missing and we are left to work mostly (almost everything) on our own without any guidance. Therefore, I ask here to see my omissions, because we all have different background. What are the valid replacements here? – PinkyWay Jan 15 '20 at 05:55
  • But still you must have read something related to evaluation of limits in your textbook or from your teacher. Try to use that. Such ad hoc procedure as in your question does not help at all. – Paramanand Singh Jan 15 '20 at 05:57
  • @ParamanandSingh, what I tried was using the commutativity of limits & continuous functions and elementary limits: $$\lim_{t\to 0}\frac{e^t-1}{t};&;\lim_{t\to 0}\frac{1-cos t}{t^2}$$ – PinkyWay Jan 15 '20 at 06:04
  • @ParamanandSingh, our exam is coming soon, and we haven't even reached that. Most of the time, I try to learn from various sources on my own and, as you see, my understanding is disputable. – PinkyWay Jan 15 '20 at 06:06
  • I am aware my mistakes are cardinal, but I thought it was better to ask on time so that all the information would accumulate. – PinkyWay Jan 15 '20 at 06:13

4 Answers4

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

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

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

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

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

lab bhattacharjee
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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 $$1\cdot 1\cdot 2\cdot 1 +0\cdot\frac{1}{2}=2$$

Paramanand Singh
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  • thank you very much! I had some of them memorised, but I would never thought of applying: $\displaystyle\lim_{x\to 0}\frac{\cosh x - 1}{x^2}$ I hope I will understand your blog some day! – PinkyWay Jan 15 '20 at 06:30
  • @VerkhovtsevaKatya: the well known standard limits are designed to handle each kind of function like logarithmic, exponential, trigonometric, hyperbolic ($\sinh, \cosh$ etc) and algebraic. Do you know the limit related to algebraic functions? – Paramanand Singh Jan 15 '20 at 06:32
  • I probably don't know most of them, apart from what I've found so far on MSE or in some books. The problem is our obligatory literature doesn't consist of typical workbooks- those are expert guides for mature mathematicians. The only way we can find something is by looking in the table of contents. All the units are totally mixed from a beginner's perspective. – PinkyWay Jan 15 '20 at 07:16
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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,

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

Quanto
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  • Your first equation is actually false because both limits are $0$. Just because the limit of two expressions are same it does not mean that one can replace one with another while evaluating limit of a complicated expression. – Paramanand Singh Jan 15 '20 at 05:56