The value of f(0) so that the function f(x) =1+x−1+x3x becomes continuous is equal to

f(x)=1x(1+x)1/2−(1+x)1/3=1x1+12x+0x2−1+13x+0x22=1x16x+0x2=16+0(x)limx→0 f(x)=16.

The value of f(0) so that the function f(x) =1+x−1+x3x becomes continuous is equal to

f(x)=1x(1+x)1/2−(1+x)1/3=1x1+12x+0x2−1+13x+0x22=1x16x+0x2=16+0(x)limx→0 f(x)=16.

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The value of $f (0)$ so that the function $f (x) =$
$\frac{\sqrt{1 + x} - \sqrt[3]{1 + x}}{x}$ becomes continuous is equal to

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answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{x} [(1 + x)^{1 / 2} - (1 + x)^{1 / 3}] \\ = \frac{1}{x} [(1 + (\frac{1}{2}) x + 0 (x^{2})) - {(1 + (\frac{1}{3}) x + 0 (x^{2}))}^{2}] \\ = \frac{1}{x} [(\frac{1}{6}) x + 0 (x^{2})] = \frac{1}{6} + 0 (x) \\ lim_{x \to 0} f (x) = \frac{1}{6} . \end{array}$