limit

u r welcome!! [1]

oops! yes,,now edited the 2nd prob!!

soln!!

\lim_{x\rightarrow\propto}\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\lim_{x\rightarrow\propto}\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}=\lim_{x\rightarrow\propto}\sqrt{\frac{x+\sqrt{x+\sqrt{x}}-\sqrt{x+\sqrt{x}}}{x+\sqrt{x+\sqrt{x}}}}=\lim_{x\rightarrow\propto}\sqrt{1-\frac{\sqrt{x+\sqrt{x}}}{x+\sqrt{x+\sqrt{x}}}}=\lim_{x\rightarrow\propto}\sqrt{1-\frac{\sqrt{\frac{1}{x^{\frac{1}{2}}}+\frac{1}{x^{\frac{3}{2}}}}}{1+\sqrt{\frac{1}{x^{\frac{1}{2}}}+\frac{1}{x^\frac{3}{2}}}}}=1

for 1) \lim_{x\rightarrow\propto}\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}

this is the intended proble right??

f(a) = sin² a - sin² ba² - b² = (sin a + sin b)(sin a - sin b)(a + b)(a - b) = (sin a + sin b) 2sin(a - b)2cos(a+b)22.(a - b)2
=(sin a + sin b) sin(a - b)2cos(a+b)2(a+b)(a - b)2

in the limit a→b

f(a→b) = sin bb.1.cos b = sin 2bb

the last question is::

\lim_{x\rightarrow a }\left(2-\frac{a}{x} \right)^\frac{{tan \Pi x}}{2a}

soln:: the bracket term is coming to be 1!!

so the answer must be 1!!!! [1] isn't it?

go to latex....the equation arises right click and select copy image link/url

then come to the message box and select image link and paste the copied image url by right click and paste option!!! :)