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.

$= \,\,\,$ $\dfrac{{[e]}^2}{{(1+0)}^3}$

$= \,\,\,$ $\dfrac{e^2}{1^3}$

$= \,\,\,$ $\dfrac{e^2}{1}$

$= \,\,\,$ $e^2$

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.