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Evaluate $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sqrt{9+5x+4x^2}-3}{x}$

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sqrt{9+5x+4x^2}-3}{x}$

Substitute $x$ is equal to zero to find the value of limit of the function.

$= \,\,\,$ $\dfrac{\sqrt{9+5(0)+4{(0)}^2}-3}{0}$

$= \,\,\,$ $\dfrac{\sqrt{9+0+0}-3}{0}$

$= \,\,\,$ $\dfrac{\sqrt{9}-3}{0}$

$= \,\,\,$ $\dfrac{3-3}{0}$

$= \,\,\,$ $\dfrac{0}{0}$

The value of the algebraic function as the limit $x$ approaches zero is indeterminate. So, the value of this function cannot be evaluated in this method.

Use Rationalization

The algebraic expression is developed in radical form. So, it is better to use rationalisation method which allows us the function gets simplified by multiplying radical form expression by its conjugate function

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sqrt{9+5x+4x^2}-3}{x}$ $\times$ $\dfrac{\sqrt{9+5x+4x^2}+3}{\sqrt{9+5x+4x^2}+3}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{(\sqrt{9+5x+4x^2}-3) \times (\sqrt{9+5x+4x^2}+3)}{x \times (\sqrt{9+5x+4x^2}+3)}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{{(\sqrt{9+5x+4x^2})}^2-{(3)}^2 }{x(\sqrt{9+5x+4x^2}+3)}$

Simplify the Limit of function

It is time to simplify the algebraic function to go ahead in finding the value of the algebraic function as the limit $x$ tends to zero.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{9+5x+4x^2-9}{x(\sqrt{9+5x+4x^2}+3)}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \require{cancel} \dfrac{\cancel{9}+5x+4x^2-\cancel{9}}{x(\sqrt{9+5x+4x^2}+3)}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{5x+4x^2}{x(\sqrt{9+5x+4x^2}+3)}$

Take common factors from the terms in the numerator of the algebraic function.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{x(5+4x)}{x(\sqrt{9+5x+4x^2}+3)}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \require{cancel} \dfrac{\cancel{x}(5+4x)}{\cancel{x}(\sqrt{9+5x+4x^2}+3)}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{5+4x}{\sqrt{9+5x+4x^2}+3}$

Evaluation of the function

Lastly, substitute $x$ is equal to zero to get the value of the algebraic function as the value of $x$ approaches zero.

$= \,\,\,$ $\dfrac{5+4(0)}{\sqrt{9+5(0)+4{(0)}^2}+3}$

$= \,\,\,$ $\dfrac{5+0}{\sqrt{9+0+0}+3}$

$= \,\,\,$ $\dfrac{5}{\sqrt{9}+3}$

$= \,\,\,$ $\dfrac{5}{3+3}$

$= \,\,\,$ $\dfrac{5}{6}$