Limits at infinity using epsilon-delta

We just started calculating limits at infinity using epsilon and delta so these might be very stupid questions.

(1) Prove that limit as $x$ goes to infinity of $\dfrac{2\sin x}{x+2}$ is $0$.

\begin{align*} |f(x)-0| &= |\frac{2\sin x}{x+2}| \\ &= |\frac{2}{x+2}| \\ &= \frac{2}{x+2} \\ &< \frac2x \end{align*} The calculations that come after this I understand but what I don't understand is why the $\sin x$ is gone just because $|\sin x|\leq 1$, and also why we have to work with a function $\frac 2x$ that is greater than our original function.

(2) Prove limit as $x$ goes to infinity of $\ln\left(\frac{x}{x+1}\right)$ is $0$ (hint: $g(t)=\ln(t)$ is continuous at $t=1$)

Here I have no clue what to do.

• Your equations are false as written. Maybe you meant to have $\le$ somewhere instead of $=$. Sep 18 '16 at 14:33

In your first question, the first "=" should be "$\leq$", since (as you point out) $|\sin(x)| \leq 1$.

For your second question, the hint is telling you to use the theorem that if $g(t)$ is continuous at $t=c$, and if $\lim\limits_{x \to a} f(x) = c$, then

$$\lim_{x \to a} g(f(x)) = g \left( \lim_{x \to a} f(x) \right)$$

In this particular example, you should use this fact, along with the fact that

$$\lim_{x \to \infty} \frac{x}{x+1} = 1$$

well,the definition is clear enough, no matter how small the epsilon is, there always x big enough to make the origin littler than it, to prove it, you find sth bigger than the origin, if u can still find the N(which the x>N)make the inequality make sense, the origin one is obviously enough. As for the second you can use the inequality ln(1+x)

• You begin typing an inequality involving $\ln$, but it's not in your post. Sep 18 '16 at 14:37
• ln(1+x)<x and its pretty obvious
– YJY
Sep 18 '16 at 14:43
• Perhaphs some formula or concrete inequalities would improve your post? Sep 18 '16 at 14:47