Hey guys having trouble solving this . Use l’Hopital’s rule to find:
$\displaystyle \lim_{x\to0}\frac{cosx-1}{x^2}$
any help much appreciated
Differentiate numerator and denominator
$\displaystyle \frac{d~}{dx} (cos(x) - 1 ) = -sin(x) $
$\displaystyle \frac{d}{dx} (x^2) = 2x
$
What is
$\displaystyle (\frac{-1}{2})\lim_{x\to0}\frac{sin(x)}{x}$ ?
You can do the same for finding this limit or just use the formula (which??)
1)Find derivatives separately
2) Divide the numerator's derivative by denominator's derivative
3)If on putting the limit you get a definite answer than game over!!
4) If not than go to step 1
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I strongly recommend reading examples here
The answer of this question will be -1/2 (after you follow above steps)
Adarsh
Denominator =D Numerator =N
Derivative of Numerator $\displaystyle = \frac{d~}{dx} (cos(x) - 1 ) = -sin(x)$...(Also called N'(x))
Derivative of Denominator $\displaystyle = \frac{d}{dx} (x^2) = 2x$...(Also called D'(x) )
According to the l'Hopital's rule
So using this
Your Limit now becomes
Once again applying the same rule this limit becomes
$\displaystyle (\frac{-1}{2}) lim_{x\to 0} cos(x)$
Now put the limit
$\displaystyle \implies (\frac{-1}{2}) \times (1)$