Nearly a Trigonometric Limit

Applying L'Hopital's rule, we can differentiate once and then cancel a factor of cos(x)-1 out of the numerator and denominator. Substituting 0 into the remaining expression, we have -(1+1)/1= -2.

Realising both numerator and denominator has limit approaching 0, L'Hopital rule is something you would wish to try. Applying it you should get $\frac {\sec^2 x - 1} {\cos x - 1}$ This might seem complicated to some, but think about secant and cosine: $\sec x = \frac {1} {\cos x}$ Substituting this, you should get $\frac {1-\cos^2 x} {(\cos^2 x)(\cos x -1)}$ Factorising the numerator, we can simplify the whole expression into $\frac {(1+ \cos x)(1-\cos x)} {(\cos^2 x)(\cos x -1)} = - \frac {1+\cos x}{\cos^2 x}$ which taking the limit gives you -2.

To solve, you must use L'Hopital's rule twice, leaving you with: lim(x-->0) of (2sec^2(x)(tan(x)/(-sin(x)). This equation simplifies down to lim(x-->0) of (2/(-cos^3(x)). Then substitute x=0 into this, and you get 2/(-1)^3, which equals -2

I don't know why this is level 2.

$\begin{array}{rcl} \lim_{x\to0}\frac{\tan x-x}{\sin x-x}&=&\lim_{x\to0}\frac{\sec^2x-1}{\cos x-1}\\ &=&\lim_{x\to0}\frac{2\sec^2x\tan x}{-\sin x}\\ &=&\lim_{x\to0}\frac{4\sec^2x\tan^2x+2\sec^4x}{-\cos x} &=&-2 \end{array}$

w.k.t tanx=x+x^3/3+2x^5/15+17x^7/315+...... so tanx-x= x^3/3+2x^5/15+17x^7/315+...... =x^3 (1/3+2x^2/15+17x^4/315+......) ====> (1) similarly sinx= x-x^3/3+x^5/5-x^7/7+.... so sinx-x= -x^3/3!+x^5/5!-x^7/7!+.... = x^3 (-1/3!+1/5!-1/7!+....) =====>(2) so (1)/(2) we get (1/3+2x^2/15+17x^4/315+......) / (-1/3!+1/5!-1/7!+....) wen we apply limit to it all terms wid x becomes zero... and we will remain with = (1/3) / (1/3!) so answer is '2'

L'Hopital Rule Type 1 for $\displaystyle \dfrac{0}{0}$

We have 2 use L-Hospital's rule. A simple point 2 keep in mind is to: Keep on differenciating both the numerator and denominator until the time your denominater part is a non-zero after you apply the limits. In this problem, we have 2 apply L-Hospital's rule 3 times 2 get the desired result.

Happy solving:)

It's basically a problem comes under category of 0/0 form.. So by L'Hospital rule we have to differentiate numerator and denominator separately till we get a proper value after substitution "0" in X...