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

Evaluate the expression lim⁡x→0 1+x2−1+x1+x3−1+x\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+{{x}^{2}}}- \sqrt{1+x}}{\sqrt{1+{{x}^{3}}-\sqrt{1+x}}}

Explanation

Solution

Hint: Use L’ Hopital Rule in this question, but remember to stop once the 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } form disappears. For that though, confirm if the expression has 00\dfrac{0}{0} or
∞∞\dfrac{\infty }{\infty } form first.
Let’s first of all see if the limit has a 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } form at all. For
that, we’ll simply substitute the limiting value of xx 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=0x=0 in
lim⁡x→0 1+x2−1+x1+x3−1+x\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,
=1+0−1+01+0−1+x=1−11−1=\dfrac{\sqrt{1+0}-\sqrt{1+0}}{\sqrt{1+0}-\sqrt{1+x}}=\dfrac{\sqrt{1}-\sqrt{1}}{\sqrt{1}-\sqrt{1}}
=00=\dfrac{0}{0} form
Hence, it is one of the indeterminate forms mentioned above. Since it has the 00\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, (1+x2−1+x)(1+x3−1+x)\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
(1+x2+1+x)(1+x3+1+x)\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,
((1+x2)−(1−x))(1+x3+1+x)((1+x3)−(1−x))(1+x2+1+x)\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)}
=x(x+1)(1+x3+1+x)x(x2+1)(1+x2+1+x)=\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)}
=(x+1)(1+x3+1+x)(x2+1)(1+x2+1+x)=\dfrac{(x+1)(\sqrt{1+{{x}^{3}}}+\sqrt{1+x})}{\left( {{x}^{2}}+1 \right)\left( \sqrt{1+{{x}^{2}}}+\sqrt{1+x} \right)}
Evaluating the limit as (x→0)\left( x\to 0 \right), we get
=(1)(1+0+1+0)(0+1)(1+0+1+0)=\dfrac{(1)(\sqrt{1+0}+\sqrt{1+0})}{\left( 0+1 \right)\left( \sqrt{1+0}+\sqrt{1+0} \right)}
⇒22=1\Rightarrow \dfrac{2}{2}=1
Thus, we found the limit to be =1=1 using this method. Now let’s try another method.
Here’s the alternate method of solving this question :
By plugging in 00 we get :
1+(0)2−1+01+(0)3−1+0=00\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 lim⁡x→a f(x)g(x)=lim⁡x→a f′(x)g′(x)=lim⁡x→a 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)}=\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: ddx((1+x2)12−(1+x)12)\dfrac{d}{dx}\left( {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{\left( 1+x \right)}^{\dfrac{1}{2}}} \right)
=12(1+x2)−12.2x−12(1+x)−12.1=\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}.2x-\dfrac{1}{2}{{\left( 1+x \right)}^{- \dfrac{1}{2}}}.1
=x(1+x2)12−12(1+x)12=\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1+x \right)}^{\dfrac{1}{2}}}}
………………..(i)
Denominator: ddx((1+x3)12−(1+x)12)\dfrac{d}{dx}\left( {{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}-{{\left( 1+x \right)}^{\dfrac{1}{2}}} \right)
=12(1+x3)−12.3x2−12(1+x)−12.1=\dfrac{1}{2}{{\left( 1+{{x}^{3}} \right)}^{-\dfrac{1}{2}}}.3{{x}^{2}}-\dfrac{1}{2}{{\left( 1+x \right)}^{- \dfrac{1}{2}}}.1
=3x22(1+x3)12−12(1+x)12=\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;

=lim⁡x→0 x(1+x2)12−12(1+x)123x22(1+x3)12−12(1+x)12=\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;
=0(1+(0)2)12−12(1+0)123(0)22(1+(0)3)12−12(1+0)12=\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}}}}}
=−12−12=1=\dfrac{-\dfrac{1}{2}}{-\dfrac{1}{2}}=1
∴lim⁡x→0 (1+x2−1+x1+x3−1+x)=1\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 11 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.,→differentiate (Numerator)differentiate (Denominator)\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.