Question

pls solve this limts question​

Answer 1

Given

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ {x}^{2} - 5x + 6}{ {x}^{2} - 9}$

Now Check the form so we put x = 3 on Given equation

$\tt \to \dfrac{(3)^{2} - 5 \times 3 + 6}{(3)^{2} - 9 }$

$\tt \to \: \dfrac{9 - 15 + 6}{9 - 9}$

$\tt \to \dfrac{15 - 15}{0} = \dfrac{0}{0}$

So it is 0/0 form , we use simplification method

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ {x}^{2} - 5x + 6}{ {x}^{2} - {3}^{2} }$

Using this identities

$\tt \to \: (a {}^{2} - b {}^{2} ) = (a - b)(a + b)$

we get

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ {x}^{2} - 5x + 6}{ ({x}- 3)(x + 3)}$

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ {x}^{2} - 2x - 3x + 6}{ ({x}- 3)(x + 3)}$

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ x({x} - 2)- 3(x - 2)}{ ({x}- 3)(x + 3)}$

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{( x- 3)(x - 2)}{ ({x}- 3)(x + 3)}$

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{ \cancel{( x- 3)}(x - 2)}{ \cancel{ ({x}- 3)}(x + 3)}$

$\to \displaystyle \lim _{ \tt \: x \to3 } \tt \dfrac{(x - 2)}{(x + 3)}$

$\to \displaystyle \tt \dfrac{(3 - 2)}{(3 + 3)} = \frac{1}{6}$

Answer is 1/6

Answer 2

Step-by-step explanation:

Answer:

$\displaystyle \boxed{{ \lim_{x \to 3} \frac{x^2+2x-15}{x^2-5x+6} =8}}$

Step-by-step explanation:

Given :

$\displaystyle\boxed{ \lim_{x \to 3} \frac{x^2+2x-15}{x^2-5x+6} }$

On solving it further :

$\begin{gathered}\displaystyle\boxed{ \lim_{x \to 3} \frac{x^2+2x-15}{x^2-5x+6} }\\\\\\\boxed\displaystyle{\rightarrow \lim_{x \to 3} \frac{x^2+5x-3x-15}{x^2-3x-2x+6} }\\\\\\\displaystyle{\rightarrow \lim_{x \to 3} \frac{x(x+5)-3(x+5)}{x(x-3)-2(x-3)} }\\\\\\\displaystyle{\rightarrow \lim_{x \to 3} \frac{(x-3)(x+5)}{(x-3)(x-2)} }\end{gathered}$

Here ( x - 3 ) get cancel out we get :

$\displaystyle\boxed{\rightarrow \lim_{x \to 3} \frac{(x+5)}{(x-2)} }$

Now putting x = 3

Now putting x = 3

$\begin{gathered}\displaystyle{\rightarrow \lim_{x \to 3} \frac{(x+5)}{(x-2)} }\\\\\\\displaystyle{\implies \frac{3+5}{3-2} }\\\\\\\displaystyle {\implies 8}\end{gathered}$