Question

evalute the limit x tends to 5​

Answer 1

Answer:

X=5

1/x-5 - 3/x^2-7x+10

= 1/5-5 - 3/5^2 - 7×5+10

= 1/0 - 3/ 25-35+10

= 1/0 -3/0

=infinity

Answer 2

$\sf \purple{ \huge \underline{\fbox{ \: Solution : \: \: }}}$

At x = 5 , the value of the given function takes the form 0/0

Therefore ,

$\sf \hookrightarrow \lim _{ x\to5} \bigg ( \frac{1}{x - 5} - \frac{3}{ {x}^{2} - 7x + 10 } \bigg) \\ \\ \sf \hookrightarrow \lim _{ x\to5} \bigg ( \frac{1}{x - 5} - \frac{3}{(x - 2)(x - 5)} \bigg) \\ \\ \sf \hookrightarrow \lim _{ x\to5} \bigg( \frac{(x - 2) - 3}{(x - 2)(x - 5)} \bigg) \\ \\ \sf \hookrightarrow \lim _{ x\to5} \bigg(\frac{x - 5}{ {x}^{2} - 7x + 10} \bigg)$

Now , evaluating the function at 5 , we get

$\sf \hookrightarrow \frac{5 - 5}{ {(5)}^{2} - 7(5) + 10} \\ \\ \sf \hookrightarrow \frac{0}{25 - 35 + 10} \\ \\ \sf \hookrightarrow 0$