Question

Any hints on where I could start with limx→0x²+sin(x²)xsin(x)+cos(x)−1

Answer 1

\[\lim_{x \rightarrow 0}\frac{ x²+\sin (x²) }{ x \sin (x) + \cos (x) -1 }\]

Answer 2

Do you know l'Hôpital's Rule?

Answer 3

Not yet, as that only comes in the second part of the course - but this is a review exercise for the first part exam so there should be a way to compute this without l'hôpital's rule

Answer 4

Anyone capable of doing this thing w/o l'Hôpital is a hero, imo.

Answer 5

You can separate them: \[\lim_{x \rightarrow 0} \frac{ x^2 }{ xsin(x)+\cos(x)-1 }+ \lim_{x \rightarrow 0}\frac{ \sin(x^2) }{xsin(x)+\cos(x)-1 }\]

Answer 6

Heroism attempt #1 ^^

What about the denominator?

Answer 7

Some hoo-doo with trig identities?

Answer 8

As suggested by @abb0t, for the first limit, you can write
\[\lim_{x\to 0}\frac{x^2}{x\sin x+\cos x-1}\]
\[\large \lim_{x\to 0}\frac{x}{\sin x+\frac{\cos x-1}{x}}\]
Now, I'm not sure about the validity of this next step, but I'm pretty sure because the function is continuous for all non-zero x, you can use the fact that
\[\frac{\cos x-1}{x}\to0 \text{ as } x\to0,\text{ leaving you with }\\
\lim_{x\to0}\frac{x}{\sin x}=1\]
And that takes care of the first limit. As for
\[\lim_{x\to0}\frac{\sin(x^2)}{x\sin x+\cos x-1},\]
I'll need a bit more time...

Answer 9

Whoops I accidentally closed this question. Anyways, I'm waiting to see if @SithsAndGiggles can solve the other half!

Answer 10

Actually, it looks like a mistake was made somewhere along the line... The first limit should come out to be 2...

Answer 11

Unfortunately, WolframAlpha isn't being very accommodating with the whole "not using L'Hopital's rule" thing.

Answer 12

Alright, I'm thinking of using the following now (so everything I did above should be disregarded):
\[\text{Near }x=0, \text{ we have }\\
\sin x\approx x\\
\sin^2x\approx (x)^2=x^2\\
\sin(x^2)\approx x^2\]
\[\lim_{x\to0}\frac{x^2+\sin^2x}{x\sin x+\cos x-1}\\
\lim_{x\to0}\frac{x^2+x^2}{x\sin x+\cos x-1}\\
2\lim_{x\to0}\frac{x^2}{x\sin x+\cos x-1}\\
2\lim_{x\to0}\frac{x^2}{x\sin x+\cos x-1}\cdot\frac{\cos x+1}{\cos x+1}\\
2\lim_{x\to0}\frac{x^2(\cos x+1)}{x\sin x(\cos x+1)+(\cos^2 x-1)}\\
2\lim_{x\to0}\frac{x^2(\cos x+1)}{x\sin x(\cos x+1)-\sin^2x}\\
2\lim_{x\to0}\frac{x^2(\cos x+1)}{\sin x\left(x(\cos x+1)-\sin x\right)}\]
Now, using the fact that
\[\frac{x}{\sin x}\to1 \text{ as } x\to0,\\
2\lim_{x\to0}\frac{x(\cos x+1)}{x(\cos x+1)-\sin x}\]
Since
\[\sin x\approx x,\\
2\lim_{x\to0}\frac{x(\cos x+1)}{x(\cos x+1)- x}\\
2\lim_{x\to0}\frac{\cos x+1}{\cos x+1- 1}\\
2\lim_{x\to0}\frac{\cos x+1}{\cos x}\\
2\cdot\frac{1+1}{1}\\
4\]
And as a check:
http://www.wolframalpha.com/input/?i=limit+of+%28x%5E2%2Bsin%28x%5E2%29%29%2F%28xsinx%2Bcosx-1%29+as+x+approaches+0