Math Doubts

Solve $\displaystyle \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{4}{x}\Bigg)}^{\displaystyle 3x}$

$x$ is a literal and the sum of the one and quotient of $4$ by $x$ is denoted as $1 + \dfrac{4}{x}$. It is raised to the power of $3x$.

${\Bigg(1+\dfrac{4}{x}\Bigg)}^{\displaystyle 3x}$

If $x$ tends to infinity, the value of the limit of the function is written as follows to express it in mathematical form.

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{4}{x}\Bigg)}^{\displaystyle 3x}$

01

Adjusting the function

The limit of the function is same the limit of the following binomial function.

$\displaystyle \Large \lim_{x \,\to\, \infty} \, \normalsize {\Bigg(1+\dfrac{1}{x}\Bigg)}^{\displaystyle x} \,=\, e$

So, try to change the limit of the function in the above form.

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\dfrac{x}{4}}\Bigg)}^{\displaystyle 3x}$

02

Adjusting the exponent of the function

Now, try to adjust the power of the function same as our formula.

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle 3x \times 1}$

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle 3x \times \normalsize \frac{4}{4}}$

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle 3 \times 4 \times \frac{x}{4}}$

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle 12 \times \frac{x}{4}}$

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle 12 \times \frac{x}{4}}$

Apply, power rule of exponents to transform the function as the power of an exponential term.

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg[{\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle \frac{x}{4}}\Bigg]}^{12}$

$= \,\,\, {\Bigg[ \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle \frac{x}{4}}\Bigg]}^{12}$

03

Adjusting the limit of the function

If $x \,\to\, \infty$ then $\dfrac{x}{4} \,\to\, \dfrac{\infty}{4}$. Therefore, $\dfrac{x}{4} \,\to\, \infty$. It is derived that the value of $\dfrac{x}{4}$ tends to infinity as $x$ approaches infinity. Change the limit of $x$ value to obtain the limit of the function in required form.

$= \,\,\, {\Bigg[ \displaystyle \Large \lim_{\normalsize \dfrac{x}{4} \Large \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)}^{\displaystyle \frac{x}{4}}\Bigg]}^{12}$

04

Obtaining Result

Finally, apply the limit formula and obtain the required result mathematically.

$= \,\,\, {[\,e\,]}^{12}$

$= \,\,\, e^{12}$

Therefore, it is the required solution for this limit problem in calculus mathematics.



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