Question

i dont get this one , lim x--> 5 sin(x-5)/x^(2)-7x+10

Answer 1

hint : factor denominator

Answer 2

then use the sinx/x limit

Answer 3

ok yea i factored the deno but how do we get sinx/x?

Answer 4

good question :)

plugin "t = x-5"
as x->5, t ->0

Answer 5

\(\large \lim \limits_{x \to 5} ~\frac{\sin(x-5)}{x^2-7x+10} \)

\(\large \lim \limits_{x \to 5} ~\frac{\sin(x-5)}{(x-5)(x-2)} \)

let \(t = x-5\)
as \(x \to 5, t \to 0\)

the limit becomes :

\(\large \lim \limits_{t \to 0} ~\frac{\sin(t)}{(t)(t+3)} \)

Answer 6

see if that looks okay so far :)

Answer 7

so is the solution 1/(t+3) , right ?

Answer 8

nope,

Answer 9

\(\large \lim \limits_{t \to 0} ~\frac{\sin(t)}{(t)(t+3)} \)


\(\large \lim \limits_{t \to 0} ~\left(\frac{\sin(t)}{(t)} \times \frac{1}{(t+3)}\right) \)

\(\large \left(\lim \limits_{t \to 0} ~\frac{\sin(t)}{(t)}\right) \times \left( \lim \limits_{t \to 0} ~\frac{1}{(t+3)} \right)\)

Answer 10

take the limits

Answer 11

so 1/3 right ??

Answer 12

\(\large \left(\lim \limits_{t \to 0} ~\frac{\sin(t)}{(t)}\right) \times \left( \lim \limits_{t \to 0} ~\frac{1}{(t+3)} \right)\)


\(\large \left( 1 \right) \times \left(\frac{1}{(0+3)} \right)\)

Answer 13

1/3 is \(\large \color{red}{\checkmark}\)