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Evaluate the limit:$$\lim_{x\to 1}\left(\frac{2-2x+\sin(x-1)}{x-1+\sin(x-1)}\right)^{\frac{1-x}{1-\sqrt x}}$$

The answer given is $\frac{1}{4}$.

My Attempt

$$\lim_{x\to 1}\left(\frac{-2+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin(x-1)}{x-1}}\right)^{1+\sqrt x}=\left(\frac{-2+1}{1+1}\right)^{1+1}=\frac{1}{4}$$

Let $f(x)=\left(\frac{-2+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin(x-1)}{x-1}}\right)^{1+\sqrt x}$

Now though I do get the answer but as $x\to 1$ the base of the limit is a negative quantity. So limit should not exist. But at the same time I feel that $x$ is required to approach $1$ through values which lie in the domain of the function so the values at which $f(x)$ does not exist do not count so there is no problem with answer being $\frac{1}{4}$

user
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Maverick
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1 Answers1

4

This kind of issue has been already discussed in a different case here

In the present case, with reference to the more general definition of limit, we can give a meaning to the expression considering rational exponents with odd denominator reduced to coprime factors as discuss in this other answer.

In this way we can define a deleted neighborhood at the point $x=1$ and take the limit.

user
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  • So the answer $\frac{1}{4}$ valid or not? According to this n this way we can define a deleted neighborhood at the point x=1 and take the limit statement I think $\frac{1}{4}$ is valid – mathophile Apr 06 '23 at 11:03
  • With reference to the more general definition of limit, yes the answer is $1/4$. – user Apr 06 '23 at 11:19