# Solving a limit with Taylor Series

I have trouble with the method of solving limits using Taylor series.

Consider for example these limits: $$\begin{align*} \lim_{x\to 0^-} \frac{1 + \log(1 + \sin\sqrt{x}) - e^{\sqrt{x}}}{\tan\sqrt{x} -\sin\sqrt{x}} &= i\infty\\ \lim_{x\to 0^+}\frac{1 + \log(1 + \sin\sqrt{x}) - e^{\sqrt{x}}}{\tan\sqrt{x} -\sin\sqrt{x}} &= -\infty \end{align*}$$

In the numerator there are three functions: $\ln()$, $\sin()$ and $e^x$. Now, I know that the function on the same side of the fraction line must have the same $o(x)$, but is the same for the nested function? The $\sin$ function, must have the same $o(x)$ of the other two?

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Why don't you use L'Hopital's Rule? – Peter Tamaroff Apr 2 '12 at 2:23

## 1 Answer

I'm not entirely sure where your problem is. Your idea is (presumably) to expand both numerator and denominator in taylor series about 0.

Let's suppose we are computing the limit through positive numbers. Since $\sqrt{x}$ is continuous then, we can just replace all the $\sqrt{x}$ by $x$.

Now from the series of $\tan$ and $\sin$, you find that the denominator is $x^3/2 + o(x^3)$.

I contend your problem is with finding a taylor series for $\log(1+\sin{x})$. There are various ways of doing this. It will turn out that we need the first three terms. The constant terms is obviously zero. To obtain two more terms, it would be possible to just differentiate twice and plug in zero. Alternatively (and I contend this is what your question is about), you can write $\log(1+y) = y - y^2/2 + o(y^2)$. You now want to substitute $y = x + o(x^2)$ (this is sine). This yields $\log(1+\sin{x}) = x - x^2/2 + o(x^2)$. I suppose you might be worried about this substitution step; however if you recall the definition of $o(...)$ you may easily verify this is correct.

Putting this together with the series for $e^x$ shows that the denominator is $-x^2 + o(x^2)$. You now have enough information to determine the limit.

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– Overflowh Jan 20 '12 at 18:25
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