Math, asked by ujjwal21062005, 22 days ago

solve.
\lim _ { x \rightarrow 0 } \frac { \sqrt { 1 + x ^ { 2 } } - \sqrt { 1 - x ^ { 2 } } } { x }

Answers

Answered by user0888
30

\huge\text{$\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{x}=0$}

\huge\text{\underline{\underline{Explanation}}}

If we take direct substitution, -

\text{$\bullet\ \displaystyle\lim_{x\to0}\left(\sqrt{1+x^{2}}-\sqrt{1-x^{2}}\right)=0$}

\text{$\bullet\ \displaystyle\lim_{x\to0}x=0$}

\bold{\cdots\longrightarrow Indeterminate\ form\ \dfrac{0}{0}.}

Let us resolve this problem. The problem arises from the form of the fraction.

We rationalize the numerator by use of the identity.

\bold{Polynomial\ identity}

\text{$\cdots\longrightarrow \boxed{(A+B)(A-B)=A^{2}-B^{2}}$}

\bold{Given\ limit,\ -}

\text{$\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{x}$}

\bold{By\ rationalization\ of\ the\ numerator,\ -}

\text{$=\displaystyle\lim_{x\to0}\left(\dfrac{1}{x}\cdot\dfrac{(1+x^{2})-(1-x^{2})}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)$}

\text{$=\displaystyle\lim_{x\to0}\left(\dfrac{1}{x}\cdot\dfrac{2x^{2}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)$}

\text{$=\displaystyle\lim_{x\to0}\dfrac{2x}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$}

\text{$=\dfrac{2\cdot0}{\sqrt{1+0}+\sqrt{1-0}}$}

\text{$=\dfrac{0}{2}$}

\text{$=0$}

So, -

\text{$\cdots\longrightarrow\boxed{\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{x}=0.}$}

\huge\text{\underline{\underline{Helpful guide}}}

*You can view the graph at \bold{desmos\ graphing\ calculator}.

We see that values near x=0 gets closer and closer to 1, but is discontinuous at x=1.

We found the limiting value because we defined a new continuous function that is defined at x=1, which we can find by substitution.

Answered by powersurya850
7

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