# Taylor limits at infinity

How can I evaluate this limit using Taylor series?

$$\lim_{x \to +\infty} \frac{x - \sin(x) \log(1+x)}{x^7}$$

I know how to evaluate the limit at $0$, but I don't know how to solve it at infinity. If this can't be solved with Taylor series, is there another "easy" way to solve it? (I know you can apply L'Hospital $7$ times, but...)

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We know that $\log(1+x)\sim_\infty \log x$ and $\log x=_\infty o(x)$ and $|\sin x|\le 1$ so $$\lim_{x\to\infty}\frac{\sin x\log (1+x)}{x}=0$$

Can you take it from here?

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Start by squeezing for positive $x$

$$\left\vert \frac{x-\sin (x) \log(1+x)}{x^7} \right\vert \leq \frac{|x|+|\sin x| |\log(1+x)|}{|x|^7}=\frac{x+|\sin x| \log(1+x)}{x^7}$$

Now, since the Maclaurin series for $\sin(x),\log(1+x)$, $$\sin(x)=x-\frac{1}{3!}x^3 +- \dots \\ \log(1+x)=x-\frac{1}{2} x^2+- \dots$$

are alternating series whose terms of nonincreasing magnitude, we have the inequalities

$$|\sin x| \leq x , \\ \log(1+x) \leq x.$$

EDIT:

As a matter of fact the second inequality holds for arbitrarily large $x$ since $\log(1+x)$ as a concave function always lies below its tangent line at $x=0$ (this line is given by the equation $y=x$).

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Don't you worry to use these expansions when $x$ goes to $\infty$ ? –  Claude Leibovici Jun 15 '14 at 8:43
@ClaudeLeibovici Thank you for your remark. I hope it's OK now. –  user1337 Jun 15 '14 at 8:53
Now, it is very good ! Cheers :) –  Claude Leibovici Jun 15 '14 at 8:56

You can use something called the Squeeze Theorem.

You know that $|\sin(x)| \leq 1$, so you know that:

$$\forall x \in \mathbb{R}, \frac{x - \log(1+x)}{x^7}\leq \frac{x - \sin(x)\log(1+x)}{x^7} \leq\frac{x + \log(1+x)}{x^7}$$

If you're willing to accept that the limit of the left expression tends to $0$ from below $0$ and the right expression tends to $0$ from above $0$, you can conclude that the middle expression, always being between these two, also tends to $0$.

$$\lim_{x \to +\infty} \frac{x - \sin(x) \log(1+x)}{x^7}=0$$

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