Question

If lim x tends to 3 px^2 -q/x-3 = 6 then find the value of p and q.
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Answer 1

Solution :

Given expression,

$\lim \: \: \: \: \: \: \: \: \tt{ \dfrac{ {px}^{2} - q }{x - 3} = 6 } \\ \tt{ x \longrightarrow \: 3}$

At 3,the given function would be in 0/0 form

L'Hospital's Rule

• Finding the derivatives of numerator and denominator and then equating the expressions to the precursor step

• Can only be used when the function tends to be in 0/0 or ∞/∞ form

Numerator

$\boxed { \begin{minipage}{5 cm} \sf{y' = \dfrac{d( {px}^{2} - q)}{dx} } \\ \\ \longmapsto \: \sf{y' = 2px \: } \end{minipage}}$

Denominator

$\boxed{\begin{minipage}{5 cm} \ \sf{y' = \dfrac{d(x - 3)}{dx} } \\ \\ \longmapsto \: \sf{y' = 1} \end{minipage}}$

• Derivative of a constant is zero

• Derivative of x w.r.t to x is one

Using L'Hospital's Rule ,

$\colon \implies \: \tt{ \lim \ \: \: \: \: \: \: \: \: 2px = 6} \\ \: \: \: \: \: \: \: \: \: \: \: \tt{x \longrightarrow \: 3}$

$\colon \implies \: \tt{6p \: = 6 } \\ \\ \colon \: \implies \: \tt{p = \dfrac{6}{6} } \\ \\ \colon \implies \: \boxed{ \boxed{ \tt{p = 1}}}$

Also,

$\lim \: \: \: \: \: \: \: \: \: \: \: \tt{ \frac{ {px}^{2} - q}{x - 3} = 6} \\ \tt{x \longrightarrow \: 3} \\ \\ \colon \: \implies \tt{ \lim \ \: \: \: \: \: \: \: \: px {}^{2} - q = 6(x - 3) } \\ \tt{ \: \: \: \: \: \: \: \: \: \: \: \: x \longrightarrow \: 3} \\ \\ \colon \implies \tt{9p - q = 0 - - - (1)}$

Putting p = 1 in equation (1),

$\leadsto \: \sf{9(1) - q= 0} \\ \\ \leadsto \: \boxed{ \boxed{ \sf{q = 9}}}$

Thus,p = 1 and q = 9