Evaluate the expression $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+{{x}^{2}}}-

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Hint: Use L’ Hopital Rule in this question, but remember to stop once the $\dfrac{0}{0}$ or $\dfrac{\infty
}{\infty }$ form disappears. For that though, confirm if the expression has $\dfrac{0}{0}$ or
$\dfrac{\infty }{\infty }$ form first.
Let’s first of all see if the limit has a $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form at all. For
that, we’ll simply substitute the limiting value of $x$ that is given to us, and then see if we’re getting and
indeterminate form like the ones mentioned above. Thus, we need to substitute $x=0$ in
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+{{x}^{2}}}-\sqrt{1+x}}{\sqrt{1+{{x}^{3}}}-\sqrt{1+x}}$.
Doing so, we get,
$=\dfrac{0}{0}$ form
Hence, it is one of the indeterminate forms mentioned above. Since it has the $\dfrac{0}{0}$ form, we
can now find it by using L’ Hopital Rule. However, let’s explore some other methods too.
For the first method, let’s follow the following path.
Take, $\dfrac{\left( \sqrt{1+{{x}^{2}}}-\sqrt{1+x} \right)}{\left( \sqrt{1+{{x}^{3}}}-\sqrt{1+x} \right)}$
Remove the subtraction of radicals by multiplying numerator and denominator by
$\left( \sqrt{1+{{x}^{2}}}+\sqrt{1+x} \right)\left( \sqrt{1+{{x}^{3}}}+\sqrt{1+x} \right)$
(That is the product of the conjugates of the numerator and the denominator).

Doing so, we get,
$\dfrac{\left( \left( 1+{{x}^{2}} \right)-\left( 1-x \right) \right)\left( \sqrt{1+{{x}^{3}}}+\sqrt{1+x}
\right)}{\left( \left( 1+{{x}^{3}} \right)-\left( 1-x \right) \right)\left( \sqrt{1+{{x}^{2}}}+\sqrt{1+x} \right)}$
$=\dfrac{x\left( x+1 \right)\left( \sqrt{1+{{x}^{3}}}+\sqrt{1+x} \right)}{x\left( {{x}^{2}}+1 \right)\left(
\sqrt{1+{{x}^{2}}}+\sqrt{1+x} \right)}$
$=\dfrac{(x+1)(\sqrt{1+{{x}^{3}}}+\sqrt{1+x})}{\left( {{x}^{2}}+1 \right)\left( \sqrt{1+{{x}^{2}}}+\sqrt{1+x}
Evaluating the limit as $\left( x\to 0 \right)$, we get
$=\dfrac{(1)(\sqrt{1+0}+\sqrt{1+0})}{\left( 0+1 \right)\left( \sqrt{1+0}+\sqrt{1+0} \right)}$
$\Rightarrow \dfrac{2}{2}=1$
Thus, we found the limit to be $=1$ using this method. Now let’s try another method.
Here’s the alternate method of solving this question :
By plugging in $0$ we get :
$\dfrac{\sqrt{1+{{\left( 0 \right)}^{2}}}-\sqrt{1+0}}{\sqrt{1+{{\left( 0 \right)}^{3}}}-
\sqrt{1+0}}=\dfrac{0}{0}$ form
Let’s use L’ Hopital Rule to solve this. L’ Hopital rule says that $\underset{x\to a}{\mathop{\lim
}}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}=\underset{x\to
a}{\mathop{\lim }}\,\dfrac{f''(x)}{g''(x)}=.....$ till our limit loses the indeterminate form. We keep
differentiating the numerator and denominator separately, until the indeterminate form goes away.
Thus, using L – HOPITAL RULE once,
We’ll differentiate numerator and denominator.
Numerator: $\dfrac{d}{dx}\left( {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{\left( 1+x
\right)}^{\dfrac{1}{2}}} \right)$
$=\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}.2x-\dfrac{1}{2}{{\left( 1+x \right)}^{-
$=\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+x \right)}^{\dfrac{1}{2}}}}$
Denominator: $\dfrac{d}{dx}\left( {{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}-{{\left( 1+x
\right)}^{\dfrac{1}{2}}} \right)$
$=\dfrac{1}{2}{{\left( 1+{{x}^{3}} \right)}^{-\dfrac{1}{2}}}.3{{x}^{2}}-\dfrac{1}{2}{{\left( 1+x \right)}^{-
$=\dfrac{3{{x}^{2}}}{2{{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+x
\right)}^{\dfrac{1}{2}}}}$ …………..(ii)
Now, divide both the equation (i) and (ii) we get;

$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}-
\dfrac{1}{2{{\left( 1+x \right)}^{\dfrac{1}{2}}}}}{\dfrac{3{{x}^{2}}}{2{{\left( 1+{{x}^{3}}
\right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+x \right)}^{\dfrac{1}{2}}}}}$
By plugging in zero;
$=\dfrac{\dfrac{0}{{{\left( 1+{{\left( 0 \right)}^{2}} \right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+0
\right)}^{\dfrac{1}{2}}}}}{\dfrac{3{{\left( 0 \right)}^{2}}}{2{{\left( 1+{{\left( 0 \right)}^{3}}
\right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+0 \right)}^{\dfrac{1}{2}}}}}$
$\therefore \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\sqrt{1+{{x}^{2}}}-
\sqrt{1+x}}{\sqrt{1+{{x}^{3}}}-\sqrt{1+x}} \right)=1$
Therefore, we get the required limit as $1$ from both the methods applied.
Note: In this type of question you can see the form by putting limits and then apply the L – HOSPITAL
RULE, that says
i.e.,$\to \dfrac{\text{differentiate (Numerator)}}{\text{differentiate (Denominator)}}$ till the fraction
loses the indeterminate form. Be careful to not differentiate it further, you might get a zero in the
denominator, in which case the fraction will become undefined itself.