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Show that the following function is continuous at the point \( (0,0) \) \[ f(x, y)=\left\{\begin{array}{ll} \frac{\sin ^{-1}(x+2 y)}{\tan ^{-1}(2 x+4 y)}, & (x, y) \neq(0,0) \\ \frac{1}{2}, & (x, y)=(0,0) \end{array}\right. \] |
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Answer» \(\lim\limits_{{(x,y)}\to(0,0)} f(x,y) \) = \(\lim\limits_{{(x,y)}\to(0,0)}\frac {sin^{-1}(x+2y)}{tan^ {-1}(2x+4y)}\) = \(\lim\limits_{x\to0}\frac {sin^{-1}(x+2mx)}{tan^ {-1}(2x+4mx)}\)(By taking y = mx) = \(\lim\limits_{x\to0}\frac {sin^{-1}(x(1+2m))}{tan ^{-1}(x(2+4m)}\)( 0/0 type) = \(\lim\limits_{x\to0}\frac {(1+2m)(1+(x(2+4m))^2)}{(2+4m)\sqrt{1-x^2(1+2m)^2}}\)( By using D.L.H Rule) = \(\frac {(1+2m)(1+0)}{2(1+2m)(1-0)^{\frac 12}}\) ( By taking limit) = 1/2 = f (0, 0) ∴ Given function is continuous at (0, 0) |
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