Q

# Stuck here, help me understand: - Let be a differentiablefunction such thatand . Then: - Limit , continuity and differentiability - JEE Main

Let $f$ be a differentiable function such that $f'(x) = 7 - \frac{3}{4}\frac{f(x)}{x}, (x>0)$ and $f(1)\neq 4$ . Then $\lim_{x\rightarrow 0^{+}} xf\left(\frac{1}{x} \right )$:

• Option 1)

exists and equals 0

• Option 2)

does not exist

• Option 3)

exist and equals $\frac{4}{7}$

• Option 4)

exists and equals 4

Views

Indeterminate forms -

The form

$\frac{0}{0},\:\frac{\infty}{\infty },0\times \infty ,1^{\infty },0^{\infty} ,\infty ^{0}, \infty-\infty\ \:$  are known as indeterminate form means

they do not exist directly

-

L - Hospital Rule -

$In \:the\:form\:of\:\:\;\frac{0}{0}\:\:and\:\:\frac{\infty }{\infty }\:\:\:we\:differentiate\:\:\frac{N^{r}}{D^{r}}\:\:separately.$

$\Rightarrow \lim_{x\rightarrow a}\:\:\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\:\:\frac{f'(x)}{g'(x)}$

- wherein

$\lim_{x\rightarrow a}\:\:\frac{\frac{d}{dx}\:f(x)}{\frac{d}{dx}\:g(x)}$

$Where \:\:f(x)\:\:and\:\:g(x)=0$

Given $f{}'(x)=7-\frac{3}{4}\frac{f(x)}{x}, \: \: x>0$

and   $f(1)\neq 4$

$\frac{dy}{dx}+\frac{3}{4}\frac{y}{x}=7$

This is in the form of L.D.E

I.F. = $e^{\int P dx}=e^{\frac{3}{4}\int \frac{1}{x}dx}$

= $x^{\frac{3}{4}}$

Now,

$y\cdot x^{\frac{3}{4}}=\int 7\cdot x^{\frac{3}{4}} dx$

$y\cdot x^{\frac{3}{4}}=7\frac{x\frac{7}{4}}{\frac{7}{4}}+C$

$f(x)=4x+Cx^{}\frac{-3}{4}$

So,

$f(\frac{1}{x})=\frac{4}{x}+Cx^{\frac{3}{4}}$

$\lim_{x\rightarrow 0^{+}}\: x\: f(\frac{1}{x})=\lim_{x\rightarrow 0^{+}}(4+C\cdot x^{\frac{7}{4}})=4$

Option 1)

exists and equals 0

Option 2)

does not exist

Option 3)

exist and equals $\frac{4}{7}$

Option 4)

exists and equals 4

Exams
Articles
Questions