If f x -1- x 1- x - e - x thenA- lim x - fx-e-2B- lim x - fx-e-2C- lim fx-1D- lim x - fx-1

The correct options are B D x→ 0lim f(x) = x→ 0lime^1/xln(1+x) e/x =x→ 0lime^ ( 1 x/2+x^2/3 ) e/x=x→ 0lime{ e^ ( x/2+x^2/3...... ) 1 }/ ( x/2+x^2/3...... ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 1

If f x =1+ x ...

Question

If f(x) = $\frac{(1 + x)^{\frac{1}{x}} - e}{x}$ then

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Solution

The correct options are
B

D

$lim x \to 0 f (x) = lim x \to 0 \frac{e^{\frac{1}{x} ln (1 + x)} - e}{x} = lim x \to 0 \frac{e^{(1 - \frac{x}{2} + \frac{x^{2}}{3})} - e}{x} = lim x \to 0 \frac{e ⎧ ⎪ ⎨ ⎪ ⎩ e^{(\frac{- x}{2} + \frac{x^{2}}{3} . . . . . .)} - 1 ⎫ ⎪ ⎬ ⎪ ⎭}{(\frac{- x}{2} + \frac{x^{2}}{3} . . . . . .)} \times \frac{(\frac{- x}{2} + \frac{x^{2}}{3} . . . . . . .)}{x} = e (ln e) (- \frac{1}{2}) = \frac{- e}{2} < - 1 (∵ e ≅ 2.7)$

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If f(x) =(1+x)1x−ex then

The correct options are B D limx→0f(x)=limx→0e1xln(1+x)−ex=limx→0e(1−x2+x23)−ex=limx→0e⎧⎪⎨⎪⎩e(−x2+x23......)−1⎫⎪⎬⎪⎭(−x2+x23......)×(−x2+x23.......)x=e(ln e)(−12)=−e2<−1 (∵ e ≅2.7)

If f(x) = $\frac{(1 + x)^{\frac{1}{x}} - e}{x}$ then