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Find this limit  \lim _{x \rightarrow 0} f(x) where \frac{x}{2(1-\sqrt{x+1})}<f(x)<\frac{x}{1+\sqrt{x+1}}-1

Option: 1

-1


Option: 2

1


Option: 3

0


Option: 4

Cannot be determined


Answers (1)

best_answer

The Sandwich theorem also known as squeeze play theorem applicable for any three real valued functions f(x),g(x) and h(x).  that shares the same domain such that f(x)\leq g(x)\leq h(x) for \forall x  in the domain of definition states the following:

For some real value of a, if \lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x), then \lim _{x \rightarrow a} g(x)=l

The provided inequality is

\frac{x}{2(1-\sqrt{x+1})} \leq f(x) \leq \frac{x}{1+\sqrt{x+1}}-1 -----------(i)

Now, perform the following calculations to the inequality (i).

\begin{aligned} & \frac{x(1+\sqrt{x+1})}{2(1-\sqrt{x+1})(1+\sqrt{x+1})} \leq f(x) \leq \frac{x(1-\sqrt{x+1})}{(1+\sqrt{x+1})(1-\sqrt{x+1})}-1 \\ & \frac{x(1+\sqrt{x+1})}{2\left(1^2-(\sqrt{x+1})^2\right)} \leq f(x) \leq \frac{x(1-\sqrt{x+1})}{\left(1^2-(\sqrt{x+1})^2\right)}-1 \end{aligned}

\begin{aligned} & \frac{x(1+\sqrt{x+1})}{2(1-(x+1))} \leq f(x) \leq \frac{x(1-\sqrt{x+1})}{(1-(x+1))}-1 \\ & \frac{x(1+\sqrt{x+1})}{-2 x} \leq f(x) \leq \frac{x(1-\sqrt{x+1})}{-x}-1 \\ & \frac{(1+\sqrt{x+1})}{-2} \leq f(x) \leq(\sqrt{x+1}-2) \end{aligned}-----------(ii)

Apply the squeeze theorem to the inequality (ii).

\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{(1+\sqrt{x+1})}{-2}\right) \leq \lim _{x \rightarrow 0} f(x) \leq \lim _{x \rightarrow 0}(\sqrt{x+1}-2) \\ & \left(\frac{(1+\sqrt{0+1})}{-2}\right) \leq \lim _{x \rightarrow 0} f(x) \leq(\sqrt{0+1}-2) \\ & -1 \leq \lim _{x \rightarrow 0} f(x) \leq-1 \end{aligned}

Therefore, the required value of the limit is 

\lim _{x \rightarrow 0} f(x)=-1

Posted by

vinayak

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