Question

limit x tends to π/4
sinx- cos x/π/4-x​

Answer 1

Solution :

We have to evaluate

$\sf\lim _{x\to\dfrac{\pi}{4}}\:\dfrac{\sin\:x-\cos\:x}{\frac{\pi}{4}-x}$

Since 0/0 is of indeterminate form.

Then ,apply L'Hospital's Rule.

$\sf\lim _{x\:\to\dfrac{\pi}{4}}\:\dfrac{\sin\:x-\cos\:x}{\frac{\pi}{4}-x} =\lim _{x\:\to\dfrac{\pi}{4}} \dfrac{\frac{d(\sin(x)}{dx} - \frac{\cos(x)}{dx}}{ \frac{d(\frac{\pi}{4})}{dx} - \frac{dx}{dx}}$

$\sf=\lim_{x\:\to\dfrac{\pi}{4}}\:\dfrac{\cos\:x-(-\sin\:x)}{1}$

$\sf=\lim_{x\:\to\dfrac{\pi}{4}}\:\cos\:x+\sin\:x$

Now put the value of limit, then

$\sf=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}$

$\sf=\dfrac{2}{\sqrt{2}}$

$\sf=\sqrt{2}$

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About L'HOSPITAL'S Rule :

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

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