# computing $\lim_{x \to 0} \frac{\sin{2x}+\arctan{3x}+3x^2}{\ln{\left(1+3x+\sin^{2}{x}\right)}+x\cdot e^x}$

Evaluate the following limit without using L'Hospital rule: $$\lim_{x\to 0} \frac{\sin{2x}+\arctan{3x}+3x^2}{\ln{\left(1+3x+\sin^{2}{x}\right)}+x\cdot e^x}$$

My try:

Ok, divide both numerator and denominator by $x$, and then it's easy to compute the limit for denominator, since $\frac{\ln{\left(1+3x+\sin^{2}x\right)}}{x}$ approaches $3$, as $x$ goes around zero. But don't know what to do with arctan part in the numerator.

• That is a good start. So all you need is an approximation for the numerator as far as the $x^1$ term. Sep 2 '14 at 11:08

Let us consider the denominator first $${\ln{\left(1+3x+\sin^{2}{x}\right)}+x\cdot e^x}$$ You certainly know that, close to $x=0$, $\sin(x)$ is almost $x$. So, the sine term can be neglected. You also know, that close to $0$ $\log(1+y)\simeq y$, so the logarithm is almost $3x$; moreover, still close to $0$, $e^x \simeq (1+x)$; then all of that makes the denominator behaving as $4x$.

Now the numerator : $\sin(2x)\simeq 2x$;since $\tan(y)\simeq y$, then $\tan ^{-1}(y)\simeq y$ so $\tan ^{-1}(3x)\simeq 3x$ and the $x^2$ can be neglected since corresponding to the second order. So, the numerator is almost $5x$.

I am sure that you can take from here and I hope you understand the path to solution.

Clearly you are on right track and you need to continue like this: \begin{aligned}L &= \lim_{x\to 0} \frac{\sin{2x}+\arctan{3x}+3x^2}{\log{\left(1+3x+\sin^{2}{x}\right)}+x\cdot e^x}\\ &= \lim_{x\to 0} \dfrac{\dfrac{\sin 2x}{x} + \dfrac{\arctan 3x}{x} + 3x}{\dfrac{\log(1 + 3x + \sin^{2}x)}{x} + e^x}\\ &= \lim_{x\to 0} \dfrac{2\cdot\dfrac{\sin 2x}{2x} + 3\cdot\dfrac{\arctan 3x}{3x} + 3x}{\dfrac{\log(1 + 3x + \sin^{2}x)}{3x + \sin^{2}x}\cdot\dfrac{3x + \sin^{2}x}{x} + e^x}\\ &= \lim_{x\to 0} \dfrac{2\cdot\dfrac{\sin 2x}{2x} + 3\cdot\dfrac{\arctan 3x}{3x} + 3x}{\dfrac{\log(1 + 3x + \sin^{2}x)}{3x + \sin^{2}x}\cdot\left\{3 + x\cdot\left(\dfrac{\sin x}{x}\right)^{2}\right\} + e^x}\\ &= \dfrac{2\cdot 1 + 3\cdot 1 + 3\cdot 0}{1 \cdot (3 + 0\cdot 1^{2}) + 1} = \frac{5}{4}\end{aligned}

We have used the following standard limits $$\lim_{y \to 0}\frac{\sin y}{y} = 1,\, \lim_{y \to 0}\frac{\arctan y}{y} = 1,\,\lim_{y \to 0}\dfrac{\log(1 + y)}{y} = 1$$

Using Taylor series, we may approximate the numerator as follows $$\underbrace{\sin{2x}}_{2x+O(x^2)}+\underbrace{ \arctan{3x} }_{3x+O(x^2)}+3x^2 = 5x + O(x^2)$$ and the denominator $$\underbrace{ \log{\left(1+3x+\sin^{2}{x}\right)}}_{3x+O(x^2)}+\underbrace{xe^x}_{x+O(x^2)} = 4x + O(x^2)$$

Thus, the limit equals $\frac{5}{4}$.