If $$L=$$ limx→$$0$$ $$1x311+x-1+ax1-bx$$ exists, then

Question

If $$L=$$ limx→$$0$$  $$1x311+x-1+ax1-bx$$  exists, then

Infinity Learn · Accepted Answer

$$L=$$ limx→$$0$$   $$1x311+x-1+ax1+bx$$       $$=$$  limx→$$0$$  $$1+bx-1+ax1+xx3$$   limx→$$0$$ $$11+x1+bx$$       $$=$$ limx→$$0$$  $$1+bx-1+ax1+x12x3$$    $$=$$ limx→$$0$$   $$1+bx-a+ax1+12x+1212-12$$!$$x2+1212-112-23$$!$$x3+$$...$$x3=$$      limx→$$0$$   $$1+bx-1+ax1+12x-18x2+116x3+$$....$$x3=$$ limx→$$0$$   $$b-12-ax+-a2+18x2+a8-116x2x3Limit$$   exists,   if   $$b-12-a=0$$  and  $$-18+a2=0$$∴              $$a=14$$ and  $$b=34$$∴         $$L=-132$$

If $L = \lim_{x \to 0} \frac{1}{x^{3}} (\frac{1}{\sqrt{1 + x}} - \frac{1 + ax}{1 - bx})$ exists, then

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If L= limx→0 1x311+x-1+ax1-bx exists, then

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If $L = \lim_{x \to 0} \frac{1}{x^{3}} (\frac{1}{\sqrt{1 + x}} - \frac{1 + ax}{1 - bx})$ exists, then