Evaluating Limits: x Approach 0 & Beyond

[doktorwho]

Homework Statement

$\frac{e^x-1}{x}$
Evaluate the limit of the expression as x approaches 0.

Homework Equations

3. The Attempt at a Solution [/B]
The question i have is more theoretical. I was able to solve this problem by expanding the expression into the talyor polynomial at $x=0$. I found that to be the easiest way to get to solution if tou asked to find a limit as x approaches 0. But my question is this:
If x approaches some constant could i do the same thing except evaluate at $x=a$? And what about if x goes to infinity?
For example $\frac{sinx}{x}$ $=0$ as x goes to inifnity but when you try the expansion it doesnt. We can't then do that, so what would then be the easiest way?

 

[BvU]

what would then be the easiest way

$$\left | {\sin x\over x } \right | \le {\left | {1\over x } \right | }$$

 

[BvU]

If x approaches some constant could i do the same thing except evaluate at $x=a$?

Yes. It comes down to a coordinate shift: replace $x-a$ by $y\ \ \$ and take $\ \displaystyle \lim_{y\downarrow 0}$

 

[Ray Vickson]

Homework Statement

$\frac{e^x-1}{x}$
Evaluate the limit of the expression as x approaches 0.

Homework Equations

3. The Attempt at a Solution [/B]
The question i have is more theoretical. I was able to solve this problem by expanding the expression into the talyor polynomial at $x=0$. I found that to be the easiest way to get to solution if tou asked to find a limit as x approaches 0. But my question is this:
If x approaches some constant could i do the same thing except evaluate at $x=a$? And what about if x goes to infinity?
For example $\frac{sinx}{x}$ $=0$ as x goes to inifnity but when you try the expansion it doesnt. We can't then do that, so what would then be the easiest way?

Do you mean you want to evaluate $\lim_{x \to a} \sin(x)/x$ for finite $a > 0$? If so, just use "continuity": $\sin(x)$ is a continuous function, so $\lim_{x \to a} \sin(x) = \sin(a)$ for any finite $a$, and
$$\lim_{x \to a} \frac{\sin(x)}{x} = \frac{\lim_{x \to a} \sin(x)}{\lim_{x \to a} x}$$
if both limits in the numerator and denominator exist and the denominator limit is not zero.

These are general properties that you should learn because they are used extensively.

As for the limit when $x \to \infty$, the solution has already been suggested by BVU, but to expand on his/her answer: use the fact that $-1 \leq \sin(x) \leq 1$, so for $x>0$ we have $-1/x \leq \sin(x)/x \leq 1/x$. Now use the "Squeeze Theorem"; see, eg.,
https://en.wikipedia.org/wiki/Squeeze_theorem
or
http://www.sosmath.com/calculus/limcon/limcon03/limcon03.html