Can someone answer this please.
limit as x approaches pi/2 (cotx / (pi/2 - x))
Any help is appreciated thanx.
Been a while, but I think:
Note that if you try to evaluate the limit by substitution, you get 0/0, which is an indeterminate form. To evaluate limits with this indeterminate form, you can use l'Hopital's rule. Take the derivative of the numerator and the denominator and THEN evaluate by substitution.
Taking the derivative gives:
$\displaystyle \frac{-csc^{2}x}{-1}$
Evaluating this by substitution (with x = pi/2) gives the limit to be 1.
Alternatively use the definiton of the derivative at a point
$\displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$
Now rewriting our limit as
$\displaystyle -\lim_{x\to\frac{\pi}{2}}\frac{\cot(x)-\cot\bigg(\frac{\pi}{2}\bigg)}{x-\frac{\pi}{2}}$
we can do this since $\displaystyle \cot\bigg(\frac{\pi}{2}\bigg)=0$
So now wee see this is exactly the derivative at point
now letting $\displaystyle f(x)=\cot(x)$ and $\displaystyle c=\frac{\pi}{2}$
we see this is equal to $\displaystyle -f'\bigg(\frac{\pi}{2}\bigg)=\csc^2\bigg(\frac{\pi} {2}\bigg)=1$
$\displaystyle \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\cot x}{\dfrac{\pi }{2}-x}=\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin \bigg( \dfrac{\pi }{2}-x \bigg)}{\dfrac{\pi }{2}-x}\cdot \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{1}{\sin x}=1.$
As Krizalid's result yielded, the limit is +1. Mathnasium didn't cancel the negative signs, and MathStud28 needed to consider $\displaystyle \frac{\pi}{2} - x = -\left(x - \frac{\pi}{2}\right)$ in the denominator in order for the definition of a derivative approach to work.