# If $f\left( 5 \right) = 7$ and $f'\left( 5 \right) = 7$ then $\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}$ is given by

$

\left( a \right)35 \\

\left( b \right) - 35 \\

\left( c \right)28 \\

\left( d \right) - 28 \\

$

Last updated date: 27th Mar 2023

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Answer

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310.5k+ views

Hint: Use L'hopital's rule, which is generally used to find the limits to reduce it to indeterminate form.

First of all we have to check any indeterminate form $\left( {\dfrac{0}{0},\dfrac{\infty }{\infty },\infty - \infty ,{0^0},{1^\infty },{\infty ^0},0 \times \infty } \right)$ make or not. If any indeterminate form comes then we apply L'hopital's rule.

Now, put $x = 5$ in $\dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}$

$ \Rightarrow \dfrac{{5f\left( 5 \right) - 5f\left( 5 \right)}}{{5 - 5}} = \dfrac{0}{0}$ (indeterminate form)

So, we apply L'hopital's rule

In L’hopital’s rule we differentiate both numerator and denominator to reduce indeterminate form.

$

\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{\dfrac{d}{{dx}}\left( {xf\left( 5 \right) - 5f\left( x \right)} \right)}}{{\dfrac{d}{{dx}}\left( {x - 5} \right)}} \\

\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{f\left( 5 \right) - 5f'\left( x \right)}}{{1 - 0}} \\

\\

$

If $x$ tends to $5$ So, there is no indeterminate form.

Now, put $x = 5$

$ \Rightarrow f\left( 5 \right) - 5f'\left( 5 \right)$

$f\left( 5 \right) = 7$ and $f'\left( 5 \right) = 7$ given in question

$

\Rightarrow 7 - 5 \times 7 \\

\Rightarrow 7 - 35 \\

\Rightarrow - 28 \\

$

So, the correct option is $\left( d \right)$.

Note: whenever we come to these types of problems first of all we have to check any indeterminate form make or not. If any indeterminate form makes So, we can apply L'hopital's rule unless the indeterminate form reduces. If there is no indeterminate form then directly put the value of $x$.

First of all we have to check any indeterminate form $\left( {\dfrac{0}{0},\dfrac{\infty }{\infty },\infty - \infty ,{0^0},{1^\infty },{\infty ^0},0 \times \infty } \right)$ make or not. If any indeterminate form comes then we apply L'hopital's rule.

Now, put $x = 5$ in $\dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}$

$ \Rightarrow \dfrac{{5f\left( 5 \right) - 5f\left( 5 \right)}}{{5 - 5}} = \dfrac{0}{0}$ (indeterminate form)

So, we apply L'hopital's rule

In L’hopital’s rule we differentiate both numerator and denominator to reduce indeterminate form.

$

\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{\dfrac{d}{{dx}}\left( {xf\left( 5 \right) - 5f\left( x \right)} \right)}}{{\dfrac{d}{{dx}}\left( {x - 5} \right)}} \\

\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{f\left( 5 \right) - 5f'\left( x \right)}}{{1 - 0}} \\

\\

$

If $x$ tends to $5$ So, there is no indeterminate form.

Now, put $x = 5$

$ \Rightarrow f\left( 5 \right) - 5f'\left( 5 \right)$

$f\left( 5 \right) = 7$ and $f'\left( 5 \right) = 7$ given in question

$

\Rightarrow 7 - 5 \times 7 \\

\Rightarrow 7 - 35 \\

\Rightarrow - 28 \\

$

So, the correct option is $\left( d \right)$.

Note: whenever we come to these types of problems first of all we have to check any indeterminate form make or not. If any indeterminate form makes So, we can apply L'hopital's rule unless the indeterminate form reduces. If there is no indeterminate form then directly put the value of $x$.

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