Evaluating limits help?

[cokeman5]

I have been having great trouble with the following problems, I not as concerned with the problems themselves, but rather the concepts behind solving them. I have attempted for hours to solve these without a solution to show for even 1 of them. I have even looked online for tips to evaluate these but only got example problems that were obvious and simple. Please help. T_T

1.lim(x-->1) [sin(x+1)/(x-x^(-1)) + sin(x+1)/(1-x^2)]

2.lim(x-->0+) (1+x)^(1/x)

3.(e^(3x)-1)/(e^x-1)

4.lim(x-->infinite) [xsin(1/x)]

5.lim(x-->infinite) e^(-x)sine(x)

6.lim(x-->0+) x^2/(1-cos(x))

 

[pappus]

I have been having great trouble with the following problems, I not as concerned with the problems themselves, but rather the concepts behind solving them. I have attempted for hours to solve these without a solution to show for even 1 of them.

We are not so much interested in your solutions but in your work so we can see what you know already and where you need some additional informations. Without this you simply ask to do your work. Ähemmm ...!

I have even looked online for tips to evaluate these but only got example problems that were obvious and simple. Please help. T_T

...

2.lim(x-->0+) (1+x)^(1/x)

...

An example:

The additional information here is: Replace x by \(\displaystyle \displaystyle{\frac1n}\)

Then your limit becomes:

\(\displaystyle \displaystyle{\lim_{x \to 0}(1+x)^{\frac1x} = \lim_{n \to \infty}\left(1+\frac1n \right)^n}\)

The RHS of this equation should be familiar to you.

 

[Yogi]

#6 you can use L'Hopital's rule.

 

[soroban]

Hello, cokeman5!

I assume #3 is supposed to look like this: .\(\displaystyle \displaystyle \lim_{x\to0}\frac{e^{3x}-1}{e^x-1}\)

Note that the numerator is a difference of cubes.

We have: .\(\displaystyle \displaystyle \lim_{x\to0}\frac{(e^x-1)(e^{2x} + e^x + 1)}{e^x-1} \;\;=\;\;\lim_{x\to0}(e^{2x} + e^x + 1) \)

. . . . . . . . . \(\displaystyle =\;\;e^0 + e^0 + 1 \;\;=\;\;1+1+1 \;\;=\;\;3\)