Question

Guys solve it please 2 no.

Answer 1

Answer:

sinα - αcosα

Step-by-step explanation:

Given that,

$\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|$

Let x - α = h

Thus when x ⇒ α , h ⇒ 0

$\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|$

$=\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-[\alpha\sin h\cos\alpha+\alpha\cos h\sin\alpha]}{h}|\\\;\\=\lim_{h\to0}\;\;|\frac{(\alpha\sin\alpha-\alpha\cos h\sin\alpha)+(h\sin\alpha-\alpha\sin h\cos\alpha)}{h}|\\\;\\=\lim_{h\to0}\;\;|\frac{\alpha\sin\alpha(1-\cos h)}{h}+\frac{h\sin\alpha}{h}-\frac{\alpha\sin h\cos\alpha}{h}|$

Now we will evaluate left hand limit and right and limit separately

For LHL:-

$\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha$

For RHL:-

$\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha$

Thus limit  is (sinα - αcosα)

Answer 2

Answer:

The limit is

Step-by-step explanation:

(sina -acosa) is the answer for the question

Hope it helps u