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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