Question

Limits
What is the maximum value of $f(1)$ for all polynomials $f(x)$ that follows the conditions?
There exists a natural number $n$ such that -$\text{ \displaystyle\lim_{x\to\infty}\dfrac{f(x)-4x^{3}+3x^{2}}{x^{n+1}+1}=6 }$
and -
$\text{ \displaystyle\lim_{x\to0}\dfrac{f(x)}{x^{n}}=4. }$

Answer 1

$\Huge\text{14}$

$\Large\text{\underline{\underline{Crucial concepts}}}$

The limit of $\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}$ contains three cases.

$\text{ \cdots\longrightarrow Degree of f(x) > Degree of g(x)\iff\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=\pm\infty. }$

$\text{ \cdots\longrightarrow Degree of f(x) = Degree of g(x)\iff\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=\dfrac{\text{Leading coefficient of f(x) }}{\text{Leading coefficient of g(x) }} }$

$\text{ \cdots\longrightarrow Degree of f(x) < Degree of g(x)\iff\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=0. }$

For -

$\cdots\longrightarrow\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\alpha$

while -

$\cdots\longrightarrow g(x)\to0$

it results in, -

$\cdots\longrightarrow f(x)\to0.$

$\Large\text{\underline{\underline{Explanation}}}$

$\text{\underline{Step 1. Check the given conditions}}$

We know that -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{f(x)-4x^{3}+3x^{2}}{x^{n+1}+1}=6. }$

So, we know it is the condition of the equal degrees of the numerator and denominator. And we even know the leading coefficient is 6.

We know that -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{f(x)}{x^{n}}=4. }$

As $x\to0$, the denominator tends to 0 and there exists a limiting value. So the numerator tends to 0 as well.

$\text{\underline{Step 2 - (A).}}$

First, let us consider $n=1$.

The leading coefficient is 6 and the degree is $n+1=2$. We even know that $f(x)$ consists of quadratic and linear terms because, -

$\cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{f(x)}{x}=4.$

Let us consider -

$\cdots\longrightarrow f(x)-4x^{3}+3x^{2}=6x^{2}+ax.$

Then, -

$\cdots\longrightarrow f(x)=4x^{3}+3x^{2}+ax.$

So, -

$\cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{f(x)}{x}=4$

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{4x^{3}-3x^{2}+ax}{x}=4 }$

$\cdots\longrightarrow\displaystyle\lim_{x\to0}(4x^{2}-3x+a)=4$

$\cdots\longrightarrow\underline{a=4.}$

$\therefore f(x)=4x^{3}+3x^{2}+4x\iff f(1)=4+3+4=11.$

$\text{\underline{Step 2 - (B).}}$

Now, let us consider $n=2$.

The leading coefficient is 6, and the degree is $n+1=3$. We even know that $f(x)$ consists of cubic and quadratic terms because, -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{f(x)}{x^{2}}=4. }$

Let us consider, -

$\cdots\longrightarrow f(x)=10x^{3}+bx^{2}.$

Since, -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{10x^{3}+bx^{2}}{x^{2}}=4 }$

it results in, -

$\cdots\longrightarrow\underline{b=4.}$

$\therefore f(x)=10x^{3}+4x^{2}\iff f(1)=10+4=14.$

$\text{\underline{Step 2 - (C).}}$

Lastly, let us consider $n\geq3$.

Let us consider -

$\cdots\longrightarrow f(x)=6x^{n+1}+cx^{n}.$

Since, -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to0}\dfrac{6x^{n+1}+cx^{n}}{x^{n}}=4 }$

it results in, -

$\cdots\longrightarrow\underline{c=4.}$

$\therefore f(x)=6x^{n+1}+4x^{n}\iff f(1)=6+4=10.$

$\text{\underline{Step 3. Conclusion part}}$

Hence, the maximum value of $f(1)$ is 14.