1. ## i need help

hi everybody, i have some problems from calculus but i can not solve , if someone can help, i will be very glad.

1) lim cos(3x)
x→1 x²+ 2x-1

2) lim sin (x-1)
x→∞ (2+x²)

3) lim sin (x²+1)
x→∞ (x+1)

4) lim (x-1) sin (x+2)
x→ -2 x²-4

5) lim √x +x²
x→-∞ x

2. Have you been given L'Hopital's Rule yet?

3. Originally Posted by dubber
hi everybody, i have some problems from calculus but i can not solve , if someone can help, i will be very glad.

1) lim cos(3x)
x→1 x²+ 2x-1

2) lim sin (x-1)
x→∞ (2+x²)

3) lim sin (x²+1)
x→∞ (x+1)
can you use L'Hopital's?

4) lim (x-1) sin (x+2)
x→ -2 x²-4
note that this is $\lim_{x \to -2} \frac {x - 1}{x - 2} \cdot \frac {\sin (x + 2)}{x + 2}$

5) lim √x +x²
x→-∞ x
note that the highest power of x is in the numerator

4. ## whats that?

i dont know what mean that ((::

5. Isn't the answer to the first one just:

$\frac{cos3}{2}$

?

Or am I making it a bit too simple?

6. Originally Posted by Showcase_22
Isn't the answer to the first one just:

$\frac{cos3}{2}$ ?
Yes

Or am I making it a bit too simple?
No

Originally Posted by dubber
2) lim sin (x-1)
x→∞ (2+x²)

3) lim sin (x²+1)
x→∞ (x+1)
Can you use the squeeze theorem ?

5) lim √x +x²
x→-∞ x
I guess it is $\frac{\sqrt{x+x^2}}{x}$, otherwise, it doesn't make much sense...

$\implies \sqrt{x+x^2}=|x| \sqrt{1+\frac 1x}=-x \sqrt{1+\frac 1x}$ since $x \to -\infty$ (so we consider x as being negative)

7. $\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}$

$\lim_{x\rightarrow \infty}\frac{-1}{2+x^2}\leq \lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\lim_{x\rightarrow \infty}\frac{1}{2+x^2}$

$\lim_{x\rightarrow \infty} \frac {-1}{x^2}< \lim_{x\rightarrow \infty}\frac{-1}{2+x^2}\leq \lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\lim_{x\rightarrow \infty}\frac{1}{2+x^2}< \lim_{x\rightarrow \infty} \frac {1}{x^2}$

$0<\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}<0$

$\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\rightarrow 0$

There's the second one. I applied the sandwich rule and got it working out. I haven't actually studied these in class so i'm not sure if that's the correct way to write it out.

I think the same method can be used for the last one. Try it and see!

8. Originally Posted by Showcase_22
$\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}$

$\lim_{x\rightarrow \infty}\frac{-1}{2+x^2}\leq \lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\lim_{x\rightarrow \infty}\frac{1}{2+x^2}$

$\lim_{x\rightarrow \infty} \frac {-1}{x^2}< \lim_{x\rightarrow \infty}\frac{-1}{2+x^2}\leq \lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\lim_{x\rightarrow \infty}\frac{1}{2+x^2}< \lim_{x\rightarrow \infty} \frac {1}{x^2}$

$0<\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}<0$

$\lim_{x\rightarrow \infty}\frac {sin(x-1)}{2+x^2}\rightarrow 0$

There's the second one. I applied the sandwich rule and got it working out. I haven't actually studied these in class so i'm not sure if that's the correct way to write it out.

I think the same method can be used for the last one. Try it and see!
the $\sin$ is a bounded function, hence if the denominator grows very much, this means $\frac{\sin(x-1)}{2+x^2}$ shrinks to $0$.
Note that $x\to\infty$ means the denominator ( $2+x^{2}$) tends to $\infty$.

Note. In general, let $f,g:[0,\infty)\to\mathbb{R}$ be two functions such that $\lim\nolimits_{t\to\infty}f(t)=0$ and $g$ is bounded.
Then, we have $\lim\nolimits_{t\to\infty}[f(t)g(t)]=0$. $\rule{0.3cm}{0.3cm}$

So, with this note, you can solve 3) and 4) too.

9. Did I write it out correctly?

I was not sure if I was supposed to write in the limit part when I was using inequalities.

10. Originally Posted by Showcase_22
Did I write it out correctly?

I was not sure if I was supposed to write in the limit part when I was using inequalities.
if you use $\lim$, the strict inequalities ( $<$ or $>$) turn to weak inequalities ( $\leq$ or $\geq$).

11. Originally Posted by Showcase_22
Did I write it out correctly?

I was not sure if I was supposed to write in the limit part when I was using inequalities.
I would have written this way :

$-1 \le \sin(x-1) \le 1$ for all x.

$\implies \frac{-1}{2+x^2} \le \frac{\sin(x-1)}{2+x^2} \le \frac{-1}{2+x^2}$

Now we can apply the limits :
$\lim_{x \to \infty} \frac{-1}{2+x^2} \le \lim_{x \to \infty} \frac{\sin(x-1)}{2+x^2} \le \lim_{x \to \infty} \frac{-1}{2+x^2}$

But $\lim_{x \to \infty} \frac{-1}{2+x^2}=\lim_{x \to \infty} \frac{1}{2+x^2}=0$

So we have the following inequality :

$0 \le \lim_{x \to \infty} \frac{\sin(x-1)}{2+x^2} \le 0$

Hence $\lim_{x \to \infty} \frac{\sin(x-1)}{2+x^2}=0$

The "problem" in what you did is that you consider since the beginning the limits. But you have to make the comparison before the limit.
Another thing : in this case, with the inequality, we can say the limit equals 0. It is no need to say it tends to 0.

12. Why's that?

Surely if I write:

$\lim_{x\rightarrow \infty} \frac {1}{x^2+2}<\lim_{x \rightarrow \infty}\frac {1}{x^2}$

Then that's right since they could never equal each other? Since:

$\frac {1}{x^2}=\frac {1}{x^2+2}$

$x^2+2=x^2$

and this has no solutions.

The "problem" in what you did is that you consider since the beginning the limits. But you have to make the comparison before the limit.
Another thing : in this case, with the inequality, we can say the limit equals 0. It is no need to say it tends to 0.
Can you say that the limit or something equals a number? The limit implies that it approaches a value but never actually equals it.

13. Originally Posted by Showcase_22
Why's that?

Surely if I write:

$\lim_{x\rightarrow \infty} \frac {1}{x^2+2}<\lim_{x \rightarrow \infty}\frac {1}{x^2}$

Then that's right since they could never equal each other? Since:

$\frac {1}{x^2}=\frac {1}{x^2+2}$

$x^2+2=x^2$

and this has no solutions.

Can you say that the limit or something equals a number? The limit implies that it approaches a value but never actually equals it.
Limits are numbers (in this case). The limit equals a number, it is the value the function approaches when x goes to infinity.

If I write $\lim_{x \to 1} x^2$, what is it ?

Yes, you surely have $\frac{1}{2+x^2} < \frac{1}{x^2}$ for any x. But their limits are equal. This is why you have to consider the week inequalities.

Another point : I don't think there is any need to introduce $\frac{1}{x^2}$ and $\frac{-1}{x^2}$. The limits of $\frac{\pm 1}{2+x^2}$ are well-known

Edit : please don't take the bold and underlined parts badly, I'm just trying to stress out the important points

14. Originally Posted by Showcase_22
Why's that?

Surely if I write:

$\lim_{x\rightarrow \infty} \frac {1}{x^2+2}<\lim_{x \rightarrow \infty}\frac {1}{x^2}$

Then that's right since they could never equal each other? Since:

$\frac {1}{x^2}=\frac {1}{x^2+2}$

$x^2+2=x^2$

and this has no solutions.

Can you say that the limit or something equals a number? The limit implies that it approaches a value but never actually equals it.
What about constant functions? It may also reach it!

Note that, we have $\frac{1}{x^{2}+2}<\frac{1}{x^{2}}$ for all $x\in\mathbb{R}$.
But dont forget that $\lim\limits_{x\to\infty}\frac{1}{x^{2}+2}=0=\lim\l imits_{x\to\infty}\frac{1}{x^{2}}$.
I always say this 'do not sanctify variables!'
Consider the followings, which are unused:
$\lim\limits_{x\to\infty}\frac{1}{x^{2}+2}\stackrel {x\mapsto\sqrt{t^{2}-2}}{=}\lim\limits_{\sqrt{t^{2}-2}\to\infty}\frac{1}{t^{2}}=\lim\limits_{t\to\inft y}\frac{1}{t^{2}}\stackrel{t\mapsto x}{=}\lim\limits_{x\to\infty}\frac{1}{x^{2}}$, note that $\sqrt{t^{2}-2}\to\infty$ as $t\to\infty$.

15. If I write , $\lim_{x\rightarrow 1}x^2$what is it ?
I would say one. I was thinking that the domain of the function $f(x)=x^2$ would include every number except 1 (ie. domain <1). So when $x \rightarrow \infty$ the value would get ever closer to one and not reach it. However, I can see why you put an equals sign. In differentiation when you differentiate from first principles, there is a value at the end eg.

$f(x)=x^2$

$\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-x^2}{h}$

$\frac{f(x+h)-f(x)}{h}=\frac {x^2+2hx+h^2-x^2}{h}$

$\frac {f(x+h)-f(x)}{h}=\frac {2hx+h^2}{h}=2x+h$

As $h \rightarrow 0$ this becomes 2x. I write in all my working $\frac {d}{dx} x^2=2x$ and ignore the fact that the tiny h exists at the end.

Yes, you surely have for any x. But their limits are equal. This is why you have to consider the week inequalities.
I understand this bit to. If x could equal then the limits would be the same. Once again, thinking about differentiation, at an infinitely small value 2x would be a tangent of x^2 ( I know I am thinking of infinitely large quantities by thinking of infinitely small ones, but I think it's helping me).

For infinite sequences I also write $S_n=something$ and not "approximately equal". In the case of infinte sequences $S_n$ is just another limit right?

Another point : I don't think there is any need to introduce and . The limits of are well-known
I just did that because my textbook does. I think it does that for sequences as well.

Edit : please don't take the bold and underlined parts badly, I'm just trying to stress out the important points
It's alright. The more I learn now the less I have to learn later.

What about constant functions? It may also reach it!
Touche! I never thought of those! I will now think of limits equalling the value a variable is approaching. Thankyou!

Thanks bkarpuz and Moo. I think I get it. When I have these lectures at university everyone is going to think i'm so great!