If f (x )= e^x^2- cos ⁡xx^2, for x 0 is continuous at x=0, then value of f (0 ) is

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Accepted Answer

Correct answer: D. (d) : Given, f (x )= e^x^2- cos ⁡xx^2 is continuous at x=0 ⇒ L.H.L. = R.H.L =f (0 ) at x=0 L.H.L. (at x=0 )= lim_h 0 0.05em e^ (0-h )^2- cos ⁡ (0-h ) (0-h )^2 = lim_h 0 0.05em e^h^2- cos ⁡h-1+1h^2= lim_h 0 0.05em( e^h^2-1h^2+ 1- cos ⁡hh^2) = lim_h 0 0.05em e^h^2-1h^2+ lim_h 0 0.05em 2 sin^2 ⁡ ( h2 )h^2 =1+2 lim_h 0 0.05em sin^2 ⁡ (h/2 ) (h/2 )^2 14=1+ 2 14= 32 Hence, f (0 )= 32

$If f (x) = \frac{e ^{x^{2}} - cos x}{x ^{2}}, for x \neq = 0 is continuous at x = 0, then value of f (0) is$

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