Limits

Question

Please give the answer with full solution.

Comments

M getting ans 2log2

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$2\log 2 = \log 2^2$

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log4,the solution provided below is right

Answer 1

$ \lim_{x\rightarrow 0} \frac{x 2^x - x}{1-\cos x}$

It forms $\frac{0}{0}$, Apply L' Hospital's Rule 

$ \lim_{x\rightarrow 0} \frac{x 2^x\log 2 + 2^x- 1}{0+\sin x}$

Again $\frac{0}{0}$, Apply L'Hospital's rule  one more time 

$ \lim_{x\rightarrow 0} \frac{(x 2^x\log 2 +2^x) \log 2 + 2^x\log 2- 0}{\cos x}$

$\\ = 2\log 2 = \log 2^2= \log 4$

Comments

Ohh..i missed the log property..hv 1 more doubt...can we apply L Hospital rule to limit in the form 1.infinity and x tending to 0?

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we apply L Hospital rule to limit in the form when we get indefinite form like 0/0 , 1/0 , infinite / infinite etc.

Answer 2

$\lim_{x\rightarrow 0} \frac{x 2^x - x}{1-\cos x}$
=  $\lim_{x\rightarrow 0} \frac{x (2^x - 1)}{2sin^{2}\frac{x}{2}}$
= $\lim_{x\rightarrow 0} \frac{2(2^x - 1)}{2sin\frac{x}{2}} * \lim_{x\rightarrow 0} \frac{x/2}{sin \frac{x}{2}}$
= $\lim_{x\rightarrow 0} \frac{2(2^x - 1)}{2sin\frac{x}{2}} * 1$ 
= $\lim_{x\rightarrow 0} \frac{2(2^x log 2)}{cos\frac{x}{2}}$ (By L hospital rule)
= $log 4$

Comments

How  $\lim_{x\rightarrow 0} \frac{x (2^x - 1)}{2sin^{2}\frac{x}{2}}$=  $\lim_{x\rightarrow 0} \frac{(2^x - 1)}{2sin\frac{x}{2}} * 1$    ???

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answer has been edited. Please tell me if approach is not correct Thanx.

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now it is correct.