EM: Limits

Question

Please solve the following question:

Comments

I am getting a=1 and b can be any real number except 0

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given answer is (D), explanation please?

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it is simple question on 1^infinity  simply apply its formula you will get the answer

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@akb1115 why b $\neq$ 0

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because while solving it we get b/x  and if we put x as infinity and if b=0 then 0/infinity is undefined

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0 / infinity is not undefined..

0 / infinity---> 0

reason is simple, you are dividing zero with very large number, it will simply be a zero.

Answer 1

This question is simply based on 1^∞ form.. which can be reduced to ∞ / ∞ form by :

Lim _{x --> ∞} (1 + f(x))^g(x)   =    e ^{lim_{x --> ∞}  f(x) g(x)} 

Alternatively if one does not remember the above formula one may take log of the given function and then may proceed.

So here :  Lim _{x --> ∞} (1 + a/x  + b/x²)^2x    =    e ^{lim_{x --> ∞}  (a/x + b/x^2) 2x}

                                                             =   e ^{lim_{x --> ∞} (a/x * 2x) + (b/x^2 * 2x)}         

                                                             =   e ^{lim_{x --> ∞} (a/x * 2x) +  lim_{x --> ∞} (b/x^2 * 2x)}  [ As lim(f(x) + g(x)) = Lim(f(x)) + Lim(g(x)) ]

As it is clear the second limit will lead to be having 2b/x , so x is in denominator and x tends to infinity so 1/x will tend to 0 , so this limit evaluates to 0 regardless of the value of the constant 'b' . Hence 'b' may be any real number.

So the overall limit                                  =  e ^{lim_{x --> ∞} (a/x * 2x)}

                                                             =  e^2a  which is given as e² in the question..

Hence a = 1 in the given limit .

Hence D) should be the correct option.

Answer 2

$\lim_{x\rightarrow \infty } (f(x))^{g(x)} = e^{\lim_{x\rightarrow \infty}g(x)(f(x) - 1)}$

Here, f(x) = $1 + \frac{a}{x} + \frac{b}{x^{2}}$ and g(x) = 2x

$$e^{\lim_{x\rightarrow \infty} 2x(1 + \frac{a}{x} + \frac{b}{x^{2}}-1)} = e^{2}$$

$\Rightarrow$ $\lim_{x\rightarrow \infty }2x\left ( \frac{a}{x} + \frac{b}{x^{2}} \right )$ = 2

$\Rightarrow$ $\lim_{x\rightarrow \infty }a + \frac{b}{x} = 1$

It is possible only when a = 1 and b any real number as $\frac{b}{x}$ will be 0 as $x\rightarrow \infty$