Question

limit x tends to pie/2 cosx/x-pie/2
PLEASE FAST

Answer 1

Solution :

We have to evaluate

$\sf\lim _{x\to\dfrac{\pi}{2}}\:\dfrac{\cos\:x}{x-\frac{\pi}{2}}$

Since 0/0 is of indeterminate form.

Then ,apply L'Hospital's Rule.

$\sf\lim _{x\:\to\dfrac{\pi}{2}}\:\dfrac{\cos\:x}{x-\frac{\pi}{2}}=\lim _{x\:\to\dfrac{\pi}{2}} \dfrac{\frac{d(\cos\:x)}{dx}}{\frac{dx}{dx}-\frac{d(\frac{\pi}{2})}{dx} }$

$\sf=\lim_{x\:\to\dfrac{\pi}{2}}\:\dfrac{-(\sin\:x)}{1}$

$\sf=\lim_{x\:\to\dfrac{\pi}{2}}\:(-\sin\:x)$

Now put the value of limit, then

$\sf=-sin(\dfrac{\pi}{2})$

$\sf=-1$

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About L'HOSPITAL'S Rule :

If f(a)=g(a)= 0 , then

$\sf \lim _{x \to \: a} \: \frac{f(x)}{g(x)} = \sf \lim _{x \to \: a} \frac{f'(x)}{g'(x)}$

Answer 2

Answer:

put x = π/2 + y

Then y = x - π/2 and y tends to 0 as x tends to π/2

lim x tends to π/2 {cos x/(x- π/2)} = lim y tends to 0 {cos (π/2 +y)/y} = lim y tends to 0 {-sin y/y} =

- lim y tends to 0 {siny/y} = - 1 Ans