Answer

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Hint: First check if the limit has an indeterminate form. If it does, only then move on to applying the L’ Hopital Rule.

The given equation is;

\[\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\sqrt{3+2x}-(\sqrt{2}+1)}{{{x}^{2}}-2}\]

Firstly, let’s check if the given limit evaluates to an indeterminate form when we put

$x=\sqrt{2}$. Let’s do that now. Doing so, we’ll have to put $x=\sqrt{2}$ in the expression

\[\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\sqrt{3+2x}-(\sqrt{2}+1)}{{{x}^{2}}-2}\]

\[\to \dfrac{\sqrt{3+2\sqrt{2}}-\left( \sqrt{2}+1 \right)}{\left( {{x}^{2}}-2 \right)}\]

\[\to \dfrac{\sqrt{3+2\sqrt{2}}-\left( \sqrt{2}+1 \right)}{0}\]

Here, we can clearly see that the denominator does equate to zero on putting the limiting value

of $x$.

Now, the numerator doesn’t look like it evaluates to zero, right ? Let’s try manipulating the

numerator alone, after we put the value $x=\sqrt{2}$ in it.

Thus, we have $\sqrt{3+2\sqrt{2}}-(\sqrt{2}+1)$

$\to \sqrt{2+1+2\sqrt{2}}-\left( \sqrt{2}+1 \right)$

$\to \sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2\sqrt{2}}-\left( \sqrt{2}+1 \right)$

The first term looks similar to the identity $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)={{\left( a+b

\right)}^{2}}$

$\to \sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}-\left( \sqrt{2}+1 \right)$

$\to{\left( \sqrt{2}+1 \right)}-{\left( \sqrt{2}+1 \right)}=0$

That becomes $0$. Hence the numerator also evaluates to zero on putting the limiting value of

$x$ in it.

Thus, we now can confirm that the limit ultimately evaluates to a $\dfrac{0}{0}$ form. Hence, it

fulfils the condition of using the L’ Hopital Rule, since the rule needs that the limit evaluates to

a $\dfrac{0}{0}$ or an $\dfrac{\infty }{\infty }$ form.

Now, by using L’HOPITAL RULE, i.e., the limit of a function evaluating to a $\dfrac{0}{0}$ or an

$\dfrac{\infty }{\infty }$ form, can be found out by separately differentiating the numerators

and denominators in terms of the variable that has the limit attached to it, successively, till the

indeterminate form goes away.

Thus, we’ll apply L’ Hopital Rule for the first time now.

Therefore, $\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\dfrac{d(\sqrt{3+2x}-

(\sqrt{2}+1))}{dx}}{\dfrac{d({{x}^{2}}-2)}{dx}}$

By power rule & chain rule, we get;

$\to \underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{1}{2\sqrt{3+2x}}.\dfrac{2-0}{2x-0}$

$\to \underset{x\to \sqrt{2}}{\mathop{\lim

}}\,\dfrac{\dfrac{1}{{2}\sqrt{3+2x}}.{2}-0}{2x-0}$

$\to \underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{1}{2x\sqrt{3+2x}}$

Now, putting the limits, we get;

$\to \dfrac{1}{2\sqrt{2}\left( \sqrt{3+2\sqrt{2}} \right)}$

$\to \dfrac{1}{2\sqrt{2}\sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2\sqrt{2}}}$

$\to \dfrac{1}{2\sqrt{2}\sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}}$

$\to \dfrac{1}{2\sqrt{2}\left( \sqrt{2}+1 \right)}$

$\to \dfrac{1}{4+2\sqrt{2}}$

Now, we will do rationalisation. Let’s multiply the numerator and denominator by the

conjugate of $4+2\sqrt{2}$. That means, we multiply numerator and denominator with $4-

2\sqrt{2}$. Doing so, we get:

$\to \dfrac{1}{\left( 4+2\sqrt{2} \right)}\times \dfrac{\left( 4-2\sqrt{2} \right)}{\left( 4-2\sqrt{2}

\right)}$

$\to \dfrac{4-2\sqrt{2}}{{{4}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}}$

$\to \dfrac{4-2\sqrt{2}}{16-8}$

$\to \dfrac{4-2\sqrt{2}}{8}$

$\to \dfrac{2-\sqrt{2}}{4}$

Hence, the solution for the equation is $\dfrac{2-\sqrt{2}}{4}$.

Note: Always check first by putting limits, if the limit evaluates to an indeterminate form, and

then, remember L-HOSPITAL RULE :

i.e., $\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 the limit

loses its indeterminate form.

The given equation is;

\[\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\sqrt{3+2x}-(\sqrt{2}+1)}{{{x}^{2}}-2}\]

Firstly, let’s check if the given limit evaluates to an indeterminate form when we put

$x=\sqrt{2}$. Let’s do that now. Doing so, we’ll have to put $x=\sqrt{2}$ in the expression

\[\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\sqrt{3+2x}-(\sqrt{2}+1)}{{{x}^{2}}-2}\]

\[\to \dfrac{\sqrt{3+2\sqrt{2}}-\left( \sqrt{2}+1 \right)}{\left( {{x}^{2}}-2 \right)}\]

\[\to \dfrac{\sqrt{3+2\sqrt{2}}-\left( \sqrt{2}+1 \right)}{0}\]

Here, we can clearly see that the denominator does equate to zero on putting the limiting value

of $x$.

Now, the numerator doesn’t look like it evaluates to zero, right ? Let’s try manipulating the

numerator alone, after we put the value $x=\sqrt{2}$ in it.

Thus, we have $\sqrt{3+2\sqrt{2}}-(\sqrt{2}+1)$

$\to \sqrt{2+1+2\sqrt{2}}-\left( \sqrt{2}+1 \right)$

$\to \sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2\sqrt{2}}-\left( \sqrt{2}+1 \right)$

The first term looks similar to the identity $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)={{\left( a+b

\right)}^{2}}$

$\to \sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}-\left( \sqrt{2}+1 \right)$

$\to{\left( \sqrt{2}+1 \right)}-{\left( \sqrt{2}+1 \right)}=0$

That becomes $0$. Hence the numerator also evaluates to zero on putting the limiting value of

$x$ in it.

Thus, we now can confirm that the limit ultimately evaluates to a $\dfrac{0}{0}$ form. Hence, it

fulfils the condition of using the L’ Hopital Rule, since the rule needs that the limit evaluates to

a $\dfrac{0}{0}$ or an $\dfrac{\infty }{\infty }$ form.

Now, by using L’HOPITAL RULE, i.e., the limit of a function evaluating to a $\dfrac{0}{0}$ or an

$\dfrac{\infty }{\infty }$ form, can be found out by separately differentiating the numerators

and denominators in terms of the variable that has the limit attached to it, successively, till the

indeterminate form goes away.

Thus, we’ll apply L’ Hopital Rule for the first time now.

Therefore, $\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{\dfrac{d(\sqrt{3+2x}-

(\sqrt{2}+1))}{dx}}{\dfrac{d({{x}^{2}}-2)}{dx}}$

By power rule & chain rule, we get;

$\to \underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{1}{2\sqrt{3+2x}}.\dfrac{2-0}{2x-0}$

$\to \underset{x\to \sqrt{2}}{\mathop{\lim

}}\,\dfrac{\dfrac{1}{{2}\sqrt{3+2x}}.{2}-0}{2x-0}$

$\to \underset{x\to \sqrt{2}}{\mathop{\lim }}\,\dfrac{1}{2x\sqrt{3+2x}}$

Now, putting the limits, we get;

$\to \dfrac{1}{2\sqrt{2}\left( \sqrt{3+2\sqrt{2}} \right)}$

$\to \dfrac{1}{2\sqrt{2}\sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}+2\sqrt{2}}}$

$\to \dfrac{1}{2\sqrt{2}\sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}}$

$\to \dfrac{1}{2\sqrt{2}\left( \sqrt{2}+1 \right)}$

$\to \dfrac{1}{4+2\sqrt{2}}$

Now, we will do rationalisation. Let’s multiply the numerator and denominator by the

conjugate of $4+2\sqrt{2}$. That means, we multiply numerator and denominator with $4-

2\sqrt{2}$. Doing so, we get:

$\to \dfrac{1}{\left( 4+2\sqrt{2} \right)}\times \dfrac{\left( 4-2\sqrt{2} \right)}{\left( 4-2\sqrt{2}

\right)}$

$\to \dfrac{4-2\sqrt{2}}{{{4}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}}$

$\to \dfrac{4-2\sqrt{2}}{16-8}$

$\to \dfrac{4-2\sqrt{2}}{8}$

$\to \dfrac{2-\sqrt{2}}{4}$

Hence, the solution for the equation is $\dfrac{2-\sqrt{2}}{4}$.

Note: Always check first by putting limits, if the limit evaluates to an indeterminate form, and

then, remember L-HOSPITAL RULE :

i.e., $\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 the limit

loses its indeterminate form.

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