Question

LIMITS: lim tends to 7 then: ((x^5)-(7^5))/((x^2)-(7^2)).​

Answer 1

$\implies \[ \lim \limits_{x\to7} \dfrac{ \sf {x}^{5} - {7}^{5} }{ \sf {x}^{2} - {7}^{2} } \]$

Now , if we put x = 7 then we will get 0/0 form :-

$\longrightarrow \: \dfrac{ \sf {x}^{5} - {7}^{5} }{ \sf {x}^{2} - {7}^{2} } \] = \dfrac{ \sf {7}^{5} - {7}^{5} }{ \sf {7}^{2} - {7}^{2} } \]$

$\longrightarrow \dfrac{ \sf {7}^{5} - {7}^{5} }{ \sf {7}^{2} - {7}^{2} } \] = \dfrac{0}{0}$

So , now we will apply L'hospital rule to solve the given problem.

We will differentiate numerator and denominator seperately.

$\implies \[ \lim \limits_{x\to7} \dfrac{ \sf5 {x}^{4} - 0 }{ \sf2 {x} - 0} \]$

$\implies \[ \lim \limits_{x\to7} \dfrac{ \sf5 {x}^{4} }{ \sf2 {x} } \]$

$\implies \dfrac{5}{2} \times \[ \lim \limits_{x\to7} { \sf {(x) }^{4 - 1} } \]$

$\implies \dfrac{5}{2} \times \[ \lim \limits_{x\to7} { \sf {x }^{3} } \]$

$\implies \dfrac{5}{2} \times {7}^{3}$

$\implies \dfrac{5}{2} \times 343$

$\implies \dfrac{1715}{2} = 857.5$