It’s basically an exponential function in terms of $x$. In this limit problem, the base and exponent both are algebraic functions in fraction form. The limit of algebraic function in exponential form has to be evaluated as $x$ approaches zero in this problem.
Let’s start simplifying this algebraic function firstly to move ahead in evaluating limit of exponential function.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{5x^2+1}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
The base of exponential function is in fractional form in which the expressions in both numerator and denominator are quadratic equations. Now, try to find the quotient of them.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+2x^2+1}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+1+2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+1}{3x^2+1}+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{\cancel{3x^2+1}}{\cancel{3x^2+1}}+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(1+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
The simplified function is similar to the following limit rule of exponential functions.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^\frac{1}{x}}$
Let’s try to transform the whole function same as the above limit rule.
Take $p = \dfrac{2x^2}{3x^2+1}$ and try to express the exponent $\dfrac{1}{x^2}$ in terms of $p$.
$\implies$ $\dfrac{3x^2+1}{2x^2} = \dfrac{1}{p}$
$\implies$ $\dfrac{3x^2+1}{x^2} = \dfrac{2}{p}$
$\implies$ $\dfrac{3x^2}{x^2}+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $\require{cancel} \dfrac{3\cancel{x^2}}{\cancel{x^2}}+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $3+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $\dfrac{1}{x^2} = \dfrac{2}{p}-3$
Therefore, the algebraic function $\dfrac{2x^2}{3x^2+1}$ can be written as $p$ simplify and the exponent $\dfrac{1}{x^2}$ can also written as $\dfrac{2}{p}-3$ in the limit of exponential function.
According to $\dfrac{1}{x^2} = \dfrac{2}{p}-3$
If $x \,\to\, 0$, then $x^2 \to {(0)}^2$. So, $x^2 \to 0$. Now $\dfrac{1}{x^2} \to \dfrac{1}{0}$, therefore $\dfrac{1}{x^2} \to \infty$ but $\dfrac{1}{x^2}$ is equal to $\dfrac{2}{p}-3$.
Therefore $\dfrac{2}{p}-3 \to \infty$. Now, $\dfrac{2}{p} \to \infty + 3$, and then $\dfrac{2}{p} \to \infty$. Finally, $\dfrac{p}{2} \to \dfrac{1}{\infty}$ and then $\dfrac{p}{2} \to 0$.
If $\dfrac{p}{2} \to 0$ then $p \to 2 \times 0$. Therefore $p \to 0$.
Therefore, if $x$ approaches $0$, then $p$ also approaches $0$.
Now, convert the limit of exponential function in terms of $x$ into its equivalent function in terms of $p$.
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(1+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$ $\,=\,$ $\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p} \,-\, 3}}$
$\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p} \,-\, 3}}$
The exponents are in subtraction form. The exponential function can be written in quotient form as per quotient rule of exponents.
$= \,\,\,$ $\displaystyle \large \lim_{p \,\to\, 0}{\normalsize \dfrac{{(1+p)}^{\frac{2}{p}}}{{(1+p)}^3}}$
According to quotient rule of limits
$= \,\,\,$ $\dfrac{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p}}}}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
According to Power of Power Rule of Exponents.
$= \,\,\,$ $\dfrac{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {{\Big[(1+p)}^{\frac{1}{p}}} \Big]}^2}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
The function in the numerator can be further simplified by the constant exponent power rule of limits.
$= \,\,\,$ $\dfrac{\displaystyle \Bigg[ \large \lim_{p \,\to\, 0}{\normalsize {{(1+p)}^{\frac{1}{p}}} \Bigg]}^2}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
The limit of the exponential function in the numerator can be evaluated by a limit rule of exponential function and the limit of the exponential function in the denominator can be evaluated by the direct substitution method.
$= \,\,\,$ $\dfrac{{[e]}^2}{{(1+0)}^3}$
$= \,\,\,$ $\dfrac{e^2}{1^3}$
$= \,\,\,$ $\dfrac{e^2}{1}$
$= \,\,\,$ $e^2$
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