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Question: Let \(\lim_{x \rightarrow \infty}\frac{ax^{2} + bx + c}{dx^{2} + ex + f}\), then the values of \(\fr...

Let limxax2+bx+cdx2+ex+f\lim_{x \rightarrow \infty}\frac{ax^{2} + bx + c}{dx^{2} + ex + f}, then the values of be\frac{b}{e} and cf\frac{c}{f} if ad\frac{a}{d} is continuous at da\frac{d}{a} are respectively

A

limx[x+x+xx]\lim_{x \rightarrow \infty}\left\lbrack \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right\rbrack

B

12\frac{1}{2}

C

log2\log 2

D

None of these

Answer

12\frac{1}{2}

Explanation

Solution

For f(x)f ( x ) to be continuous at x=0x = 0

limx0f(x)=f(0)=limx0+f(x)\lim _ { x \rightarrow 0 ^ { - } } f ( x ) = f ( 0 ) = \lim _ { x \rightarrow 0 ^ { + } } f ( x )

Now,

limx0+etan2x/tan3x=limx0+e(tan2x2x2x)/(tan3x3x3x)=limx0+e2/3=e2/3\lim _ { x \rightarrow 0 ^ { + } } e ^ { \tan 2 x / \tan 3 x } = \lim _ { x \rightarrow 0 ^ { + } } e ^ { \left( \frac { \tan 2 x } { 2 x } - 2 x \right) / \left( \frac { \tan 3 x } { 3 x } 3 x \right) } = \lim _ { x \rightarrow 0 ^ { + } } e ^ { 2 / 3 } = e ^ { 2 / 3 }

\therefore ea=b=e2/3e ^ { a } = b = e ^ { 2 / 3 }a=23a = \frac { 2 } { 3 } and b=e2/3b = e ^ { 2 / 3 }