Question

Evaluate the value of $\mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right)$

Answer 1

Hint : In this question, we need to evaluate the value of the function $\mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right)$ . For this, we will first check whether the given function is determinate or indeterminate and if found indeterminate then, we will use the L-Hospital’s rule.

Complete step by step solution:
The given function is $\mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right)$.
Substituting the value of ‘h’ in the given function, we get
$\mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right) = \dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt \infty }}{{{\infty ^{\left( {\dfrac{3}{2}} \right)}}}} \\\ = \dfrac{\infty }{\infty } \\\$
From the above equation, we can see that the value of the function is indeterminate at the infinite value of ‘h’.
So, applying L-Hospital’s rule which tells us that if we have an indeterminate form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ all we have to do is to differentiate the numerator and differentiate the denominator and then take the limit. L hospitals are applicable only if the value of f and g is 0, where f and g are defined functions.
Mathematically, applying the L-Hospital’s rule implies that we have to differentiate the numerator and the denominator individually with respect to the parameter present in the function.
Differentiating the given function with respect to the only parameter ‘h’, we get
$\Rightarrow \mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\dfrac{d}{{dh}}\left( {\sqrt 1 + \sqrt 2 + .... + \sqrt h } \right)}}{{\dfrac{d}{{dh}}\left( {{h^{\left( {\dfrac{3}{2}} \right)}}} \right)}}} \right) = \mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\left( {\dfrac{1}{{2\sqrt h }}} \right)}}{{\left( {\dfrac{{3\sqrt h }}{2}} \right)}}} \right) \\\ = \mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{1}{{3h}}} \right) \\\$
Now, substituting the value of h as infinity in the above equation

$\Rightarrow \mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right) = \mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{1}{{3h}}} \right) \\\ = \dfrac{1}{{3(\infty )}} \\\ = 0 \\\$
Hence, the value of the function $\mathop {\lim }\limits_{h \to \infty } \left( {\dfrac{{\sqrt 1 + \sqrt 2 + .... + \sqrt h }}{{{h^{\left( {\dfrac{3}{2}} \right)}}}}} \right)$ is equals to zero.

Note : If the given function is in the indeterminate form of $\dfrac{0}{0}$ then we apply the L-Hospital’s rule where we differentiate the numerator and the denominator of the function until we get a non-zero solution. We have to apply the L-Hospital’s rule on the function until we get the determinate form.