Limit with Taylor series

Question

I'm stuck computing this limit. I've already tried with Taylor series but I can't find the solution.

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

Answer 1

$$\dfrac{e^x\log(1+x)+x \sqrt{1-2x}}{\sin{x^2}}=$$ $$=\frac{\left(1+x+\frac{x^2}{2}+...\right)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-...\right)+x-x^2-\frac{x^3}{2}+...}{x^2-\frac{x^6}{6}+...}=\frac{2x+...}{x^2+..}\rightarrow\pm\infty.$$

I think you made the typo and you mean the following:

$$\dfrac{e^x\log(1+x)-x \sqrt{1-2x}}{\sin{x^2}}=$$ $$=\frac{\left(1+x+\frac{x^2}{2}+...\right)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-...\right)-x+x^2+\frac{x^3}{2}+...}{x^2-\frac{x^6}{6}+...}=\frac{\frac{3}{2}x^2+...}{x^2+..}\rightarrow\frac{3}{2}.$$

Answer 2

With L'Hospital:

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

and the limit doesn't exist finitely since the denominator above tends to to $\;1+1\;$ , whereas the denominator to zero ... In fact, the limit does not exist at all since the one sided limits are different ($\;\pm\infty\,$).

Care! The above does not mean we cannot use L'Hospital, since if we take the one sided limits then in each the generalized limit exists.

Answer 3

$$\lim_{x\to 0}\frac{e^x \log(1+x)+x \sqrt{1-2x}}{\sin(x^2)}$$

Use $\displaystyle\log(1+x) = x-\frac{x^2}{2} +\mathcal O({x^2})$

and binomial expansion $\displaystyle \sqrt{1-2x}=1-4x+\mathcal O(x)$ ; to get

\begin{align}\lim_{x\to 0}\frac{e^x \log(1+x)+x \sqrt{1-2x}}{\sin(x^2)}&=\lim_{x\to 0}\frac{e^x \left( x-\frac{x^2}{2}\right)+x \left( 1-4x\right)}{\dfrac{\sin(x^2)}{x^2} \cdot x^2} \\ &=\lim_{x\to 0}\frac{x(e^x+1) -e^x \frac{x^2}{2}-4x^2}{x^2}\\ &=\left[\lim_{x\to 0}\frac{e^x+1}{x} \right]-\left[\lim_{x\to 0} \frac{e^x}{2}+4\right]\\ &=\underbrace{\left[\lim_{x\to 0}\frac{e^x+1}{x} \right]}_{\to \infty}- \frac{9}{2}\\ \end{align}

Hence, limit doesn't exist.

Answer 4

Note that we don't need Taylor's expansion, indeed

$$\dfrac{e^x\log(1+x)+x \sqrt{1-2x}}{\sin{x^2}} =\\=\frac{x^2}{\sin x^2}\left(\frac{e^x}{x}\dfrac{\log(1+x)}{x}+\dfrac{ \sqrt{1-2x}}{x}\right)\to1\cdot(\pm\infty\cdot1\pm\infty)=\pm\infty$$

thus

$$\lim_{x\to 0^+}\dfrac{e^x\log(1+x)+x \sqrt{1-2x}}{\sin{x^2}}=+\infty$$

$$\lim_{x\to 0^-}\dfrac{e^x\log(1+x)+x \sqrt{1-2x}}{\sin{x^2}}=-\infty$$