Evaluate the limit $\lim_{x \to 0}\frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cosh x}{x(\sqrt{1+2 x}-\sqrt[3]{1+3 x})}$ Evaluate the limit $$L=\lim _{x \to 0}\frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cos h x}{x(\sqrt{1+2 x}-\sqrt[3]{1+3 x})}$$
By generalized binomial expansion we have
$$\sqrt{1+2 x}-\sqrt[3]{1+3 x}=\left[1+\frac{1}{2}(2 x)+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}(2 x)^{2}+\cdots\right]-\left[1+\frac{1}{3}(3 x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2}(3 x)^{2}+\cdots\right]$$
$\implies$
$$\sqrt{1+2x}-\sqrt[3]{1+3x}=\frac{x^2}{2}+O(x^3)$$
$\implies$
$$L=\lim _{x \rightarrow 0} \frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cos h x}{x^{3}\left(\frac{\frac{x^{2}}{2}+O\left(x^{3}\right)}{x^{2}}\right)}$$
$\implies$
$$L=2 \lim _{x \rightarrow 0} \frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\left(\frac{e^{x}+e^{-x}}{2}\right)}{x^{3}} \to (1)$$
We have $$e^{x}=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\ldots$$ and
$$\cos x=1-\frac{1}{2} x^{2}+\frac{1}{24} x^{4}-\frac{1}{720} x^{6}+.$$
Thus we have
$$e^x-\cos x=x+x^2+\frac{x^3}{6}+O(x^4)$$
Now we have the Maclaurin's series expansion of $\tan x$ as:
$$\tan x=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+...$$
So we get
$$\tan(e^x-\cos x)=x+x^2+\frac{x^3}{6}+\frac{1}{3}x^3+O(x^4) \to (2)$$
Also we have
$$e^{x}+\cosh x=e^{x}+\frac{e^{x}+e^{-x}}{2}$$
$$e^x+\cos hx=\begin{aligned}
\frac{3}{2}(1+x+&\left.\frac{x^{2}}{2!}+\frac{x^{3}}{3 !}+..\right) 
+\frac{1}{2}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2!}-\frac{x^{3}}{3 !}+\cdots^{}\right)
\end{aligned}$$
$\implies$
$$e^x+\cos hx=2+x+x^2+\frac{x^3}{6}+O(x^4) \to (3)$$
Using $(2),(3)$ in $(1)$ we get
$$L=\frac{2}{3}$$
Is there any alternate way?
 A: There is one right way to do this limit:
$$ \lim_{x \to 0} \frac{  
\frac{1- \cosh x}{x}+  \left[ \frac{\tan(e^x - \cos x)}{x} -\frac{1-e^x}{x} \right]}{
\sqrt{1 +2x}- (1+3x)^{\frac13}}=\lim_{x \to 0} \frac{\left( - \frac{x}{2!} - \frac{x^3}{4!}+O(x^5)\right)+ \left[ \frac{\left(2e^x -1- \cos x  \right)+\frac{(e^x - \cos x)^3}{3}}{x}.. \right]}{\left[ 2-1\right] \frac{x^2}{2!} +O(x^3)}$$
Note that:
$$ (2e^x - 1 - \cos x)= \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$:
Hence,
$$ \lim_{x \to 0}\frac{ \left( - \frac{x}{2!} - \frac{x^3}{4!}+O(x^5) \right)+ \left[ \frac{\frac{x^2}{2} + \frac{x^3}{3} + O(x^4) }{x}.\right]}{ \frac{x^2}{2!} +O(x^3)} $$
$$ \lim_{x \to 0}\frac{\frac{x^2}{3} + O(x^3)}{\frac{x^2}{2!}+O(x^3)}= \frac23$$
Idea: I first rearranged the expression to make them all look like derivatives, for the denominator I tried to find the first non zero term of in it's expansion and then expanded numerator in that.
It is still a mess, but it reduces solving the question in an algorithim.


*Arrange the expressions to look like known limits on numerator and denominator

*Find deg denominator

*Expand numerator till deg denominator

*Careful algebra

*Take the limit

