2

Please help calculation (and understanding) this limit:

$$ \lim_{x\to 1} \frac{x \cos (x-1)-1}{x-1} $$

Not sure if this is the way but I used t = x - 1:

$$ \lim_{t\to 0} \frac{(t+1)\cos(t)-1}{t} $$ and then: $$ \lim_{t\to 0} \frac{(t+1)\cos(t)-\sin^2(t)-\cos^2(t)}{t} $$

and more or less here I got stuck..Not even sure this is the right direction. I need more than just a solution, I need full explanation please

YukiJ
  • 2,529
Noam
  • 853
  • 4
  • 16
  • If you don't want to use the L'Hospital rule or derivative, you can use the identity cos(2x)-1=-2sin^2(x) to transform it to the well-known limit of sin(t)/t – Huang Dec 29 '16 at 23:32
  • https://math.stackexchange.com/questions/1069642/finding-a-limit-using-change-of-variable-how-come-it-works https://math.stackexchange.com/questions/1066257/calculate-a-trigonometric-limit – Workaholic Dec 30 '16 at 10:55

3 Answers3

2

Actually, you should notice that

$$f(x)=(x+1)\cos(x)$$

If you take the derivative at $x=0$, you end up with

$$f'(0)=\lim_{t\to0}\frac{(t+1)\cos(t)-1}t$$

Simply Beautiful Art
  • 74,685
1

I think it might be helpful to break the numerator up: $$\lim_{t \to 0} \frac{(t+1)\cos t-1}{t}=\lim_{t \to 0} \frac{t\cos t+\cos t-1}{t}$$ Now, we have one term with $t\cos t$ and another part with $\cos t-1$. We know that as $t \to 0$, we have $\cos t \to 1$ and thus $\cos t-1 \to 0$. Therefore, let's break the limit up. First, do the $t\cos t$ part: $$\lim_{t \to 0} \frac{t\cos t}{t}=\lim_{t \to 0} \cos t=1$$ Now, for the $\cos t-1$ part, we get $\cos t-1\to 0$ in the numerator and $t \to 0$ in the denominator, so we need to simplify further. Because we have $\cos t-1$ in the numerator, I am going to multiply by the "conjugate," which is $-\cos t-1$: $$\frac{\cos t-1}{t}\frac{-\cos t-1}{-\cos t-1}=\frac{1-\cos^2 t}{t}\frac{1}{-\cos t-1}=\frac{\sin^2 t}{t}\frac{1}{-\cos t-1}=\frac{\sin t}{t}\frac{\sin t}{-\cos t-1}$$ It is well-known (from Squeeze Theorem) that $\lim_{t \to 0} \frac{\sin t}{t}=1$. Also, $\lim_{t \to 0} \frac{\sin t}{-\cos t-1}=\frac{\sin 0}{-\cos 0-1}=0$, so the whole second part is $1\cdot 0=0$.

This gives us the whole answer, first part plus second part, to be $1+0=1$.

Noble Mushtak
  • 18,402
  • 28
  • 44
  • I get your logic but It isn't formal enough to say "..we can ignore that part.." please rephrase as formal as possible.. to me I just look at this and see (cost -1)t -> 0/0 when t->0 so I need to keep simplifying the expression.. – Noam Dec 29 '16 at 23:16
  • @Noam Good point. I have edited my answer to make it more formal. – Noble Mushtak Dec 29 '16 at 23:18
  • Thank you, the first part is perfectly clear, but as I can't use L'Hopital's Rule just yet (haven't learned it), is there any chance you can solve the second part in a different way? more like you did with the first part? – Noam Dec 29 '16 at 23:22
  • @Noam Yes, I can solve the second part differently. It can be done without calculus as shown here. I will edit this into my answer. – Noble Mushtak Dec 29 '16 at 23:22
  • You are the best. many thanks. – Noam Dec 29 '16 at 23:24
  • @Noam OK, I have used no calculus in my new answer. – Noble Mushtak Dec 29 '16 at 23:25
0

Without calculus:

Start with the geometrically obvious formula

$$\lim_{x\to 0} \, \frac{\sin (x)}{x}=1$$

Next use the triginometric identity

$$\cos (t)=\cos ^2\left(\frac{t}{2}\right)-\sin ^2\left(\frac{t}{2}\right)=1-2 \sin ^2\left(\frac{t}{2}\right)$$

Both relations together lead to the approximation

$$\cos (t)\simeq 1-\frac{t^2}{2}$$

for $t\to 0$.

Inserting this into the expression in question and simplifying elementarily you can safely let $t\to 0$ to find the limit 1.

Dr. Wolfgang Hintze
  • 12,465