Question

solve this limit value.............

Answer 1

Answer:

1/4

Step-by-step explanation:

Lim (√(x+4) -2) / x

    =(√(x+4) -2) * (√(x+4)+2) / x * (√(x+4)+2)

    =(x+4-4) / (x * (√(x+4)+2))

    = x/x * (√(x+4)+2)

    = 1/√(x+4)+2

putting x=0

   = 1/ √4 +2

   =1/4

Answer 2

Solution 1:

We need to evaluate the given limit:

$\sf\implies \lim \limits_{ x \to 0} \dfrac{ \sqrt{x + 4} - 2 }{x}$

If we directly substitute the limits, we get:

${ \sf\implies \dfrac{ \sqrt{0 + 4} - 2 }{0} = \dfrac{ \sqrt{2} - \sqrt{2} }{0} = \boxed{\dfrac{0}{0}} }$

this is an indeterminate quantity, therefore substitution method failed to evaluate our limit. Let's solve it using the method of rationalisation.

$\sf\implies \lim \limits_{ x \to 0} \dfrac{ \sqrt{x + 4} - 2 }{x} \times \dfrac{ \sqrt{x + 4} + 2 }{ \sqrt{x + 4} + 2 }$

$\sf\implies \lim \limits_{ x \to 0} \dfrac{(\sqrt{x + 4} - 2)( \sqrt{x + 4} + 2) }{x( \sqrt{x + 4} + 2) }$

$\sf\implies \lim \limits_{ x \to 0} \dfrac{(\sqrt{x + 4})^{2} - {(2)}^{2} }{x( \sqrt{x + 4} + 2) }$

$\sf\implies \lim \limits_{ x \to 0} \dfrac{{x + 4} - 4}{x( \sqrt{x + 4} + 2) }$

$\sf\implies \lim \limits_{ x \to 0} \dfrac{ \not x }{ \not x( \sqrt{x + 4} + 2) }$

$\sf\implies \lim \limits_{ x \to 0} \dfrac{ 1}{ \sqrt{x + 4} + 2 }$

Now directly substituting the limits:

$\sf\implies \dfrac{ 1}{ \sqrt{0 + 4} + 2 }$

$\sf\implies \dfrac{ 1}{2 + 2 }$

$\sf\implies \dfrac{ 1}{4}$

So the required answer is:

$\boxed{ \tt\lim \limits_{ x \to 0} \dfrac{ \sqrt{x + 4} - 2 }{x} = \dfrac{1}{4} }$

$\rule{280}{1}$

Solution 2:

We need to prove that the given limit is equal to 1/2. Let's start will LHS!

$\sf\implies \lim\limits_{x\to \infty} (\sqrt{x^2+x} - x)$

If we directly substitute the limits, we get:

$\sf{\implies (\sqrt{ \infty^2+ \infty } - \infty ) = \boxed{ \infty - \infty }}$

this is an indeterminate quantity. Therefore substitution method fails to evaluate this limit. We need to use another method.

Consider LHS:

$\sf\implies \lim\limits_{x\to \infty} (\sqrt{x^2+x} - x)$

Now, rationalise the denominator.

$\sf{\implies \lim\limits_{x\to \infty} (\sqrt{x^2+x} - x) \times \dfrac{( \sqrt{ {x}^{2} + x } + x) }{ (\sqrt{ {x}^{2} + x } + x) } }$

$\sf{\implies \lim\limits_{x\to \infty} \dfrac{(\sqrt{x^2+x} - x) ( \sqrt{ {x}^{2} + x } + x) }{ (\sqrt{ {x}^{2} + x } + x) } }$

$\sf{\implies \lim\limits_{x\to \infty} \dfrac{(\sqrt{x^2+x})^{2} - {(x)}^{2} }{ (\sqrt{ {x}^{2} + x } + x) } }$

$\sf{\implies \lim\limits_{x\to \infty} \dfrac{ \not x^2+x - \not{x}^{2} }{ \sqrt{ {x}^{2} + x } + x} }$

$\sf{\implies \lim\limits_{x\to \infty} \dfrac{ \not x}{ \not x( \sqrt{1 + \frac{1}{x} } + 1)} }$

$\sf{\implies \lim\limits_{x \to \infty} \dfrac{1}{\sqrt{1 + \frac{1}{x} } + 1} }$

Now substituting the limits:

$\sf{\implies \dfrac{1}{\sqrt{1 + \frac{1}{ \infty} } + 1} }$

$\sf{\implies \dfrac{1}{\sqrt{1 + 0} + 1} }$

$\sf{\implies \dfrac{1}{1+ 1} }$

$\sf{\implies \dfrac{1}{2} }$

Which is equal to RHS.

Hence the required result is proved that:

$\boxed{ \tt\lim\limits_{x\to \infty} (\sqrt{x^2+x} - x) = \dfrac{1}{2} }$