# Thread: [SOLVED] Limit confusion: different methods, different solutions :S

1. ## [SOLVED] Limit confusion: different methods, different solutions :S

Hey there

note: x approaches 0 on the limit, didn't know how to format the code
I was solving this problem on my textbook:
$lim((e^(x^2)-cos(x))/x^2)=?$
On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

On the other hand, I tried this solution:

From the limit laws we get:
$lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)$
we know that limcos(x) as x approaches 0 is 1, so we end up:
lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

Thanks

2. Originally Posted by FreeedomOverdose
Hey there

note: The x approaches 0 on the limit, didn't know how to format the code
I was solving this problem on my textbook:
$lim((e^(x^2)-cos(x))/x^2)=?$
On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

On the other hand, I tried this solution:

From the limit laws we get:
$lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)$
we know that limcos(x) as x approaches 0 is 1, so we end up:
lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

Thanks
Your statment "we end up: lim((e^(x^2)- 1)/x^2)" is wrong. You have already taken the limit in one part of the formula but not the others- you can't do that.

3. Hey

lim(e^(x^2)-cos(x))=lim(e^(x^2))-lim(cos(x))= lim(e^(x^2))-1
now, we replace 1 with lim(1) (as x approaches 0)
so we have:
lim(e^(x^2))-lim(1) => (sum of limits=limit of sum)lim((e^(x^2))-1)

:S

4. Originally Posted by FreeedomOverdose
Hey there

note: The x approaches 0 on the limit, didn't know how to format the code
I was solving this problem on my textbook:
$lim((e^(x^2)-cos(x))/x^2)=?$
On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

On the other hand, I tried this solution:

From the limit laws we get:
$lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)$
we know that limcos(x) as x approaches 0 is 1, so we end up:
lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

Thanks

You cannot use arithmetic of limits if you don't know that ALL the limits involved exist finitely

In the present case, $\lim_{x\to 0}\frac{e^{x^2}-\cos x}{x^2}\neq \frac{\lim_{x\to 0}e^{x^2}-\lim_{x\to 0}\cos x}{\lim_{x\to 0}x^2}$ since the limit in the DENOMINATOR equals zero...!

And after this you allow yourself to calculate ONLY the limit of the cosine but NOT the other ones....wrong.

Perhaps the simplest way is using L'Hospital's rule:

$\lim_{x\to 0}\frac{e^{x^2}-\cos x}{x^2}=\lim_{x\to 0}\frac{2xe^{x^2}+\sin x}{2x}=\lim_{x\to 0}\left[e^{x^2}+\frac{1}{2}\,\frac{\sin x}{x}\right] = 1 + \frac{1}{2}=\frac{3}{2}$

Tonio

what I mean is:
$lim(e^(x^2))-lim(cos(x))$. Since $lim(cos(x))=1$, we replace 1 with lim(1) (as it is also equal to 1), so we have:

$lim(e^(x^2))-lim(1)$, which should be equal to $lim(e^(x^2)-1)$

:|

6. ps: i think this should be listed at the pre-calculus section, sorry abt it

7. Originally Posted by FreeedomOverdose

what I mean is:
$lim(e^(x^2))-lim(cos(x))$. Since $lim(cos(x))=1$, we replace 1 with lim(1) (as it is also equal to 1), so we have:

$lim(e^(x^2))-lim(1)$, which should be equal to $lim(e^(x^2)-1)$

:|

I know that's what you meant, and if you hadn't what you have in the denominator it'd be correct, but then do it all at once: since $\lim_{x\to 0}e^{x^2}=1$ , substitute both limits by their numerical value and then you get $0$ ...but then you don't get what you wanted to, right?

tonio

8. so, the denominator is the problem?! So am I correct to say that:

lim(e^x^2-cos(x))=lim(e^x^2-1)?!

9. Originally Posted by FreeedomOverdose
so, the denominator is the problem?! So am I correct to say that:

lim(e^x^2-cos(x))=lim(e^x^2-1)?!
If $\lim_{x\to a}f(x),\lim_{x\to a}f(x)$ exist then $\lim_{x\to a}\left\{f(x)\pm g(x)\right\}=\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)$

10. thank you guys! Everything is clear now. I just got myself into a vicious circle, happens sometimes :P.