The first and last expressions are identified as a limit that exists.
'In Mathematics, a limit is defined as a value that a function approaches the output for the given input values.'
According to the given problem,
For expression 1,
[tex]\lim_{n \to4}\frac{x^{2}-x-12 }{2-x}[/tex]
⇒ [tex]\lim_{n \to4} \frac{x^{2}-4x+3x-12 }{4-x}[/tex]
⇒ [tex]\lim_{n \to4}\frac{-x(4-x)-3(4-x)}{4-x}[/tex]
⇒ [tex]\lim_{n \to4}\frac{(4-x)(-x-3)}{4-x}[/tex]
⇒ [tex]\lim_{n \to4 }(-x-3)[/tex]
Now, we put the value of x as 4
⇒ - 4 - 3
= -7
Therefore, the limit exists.
For the last expression,
[tex]\lim_{x \to 2} \frac{x^{2}+3x-10 }{x^{2} -4}[/tex]
⇒ [tex]\lim_{x \to 2} \frac{x^{2}+5x - 2x-10 }{(x+4)(x-4)}[/tex]
⇒ [tex]\lim_{x \to2} \frac{x(x+5) -2(x+5)}{(x+4)(x-4)}[/tex]
⇒ [tex]\lim_{n \to 2} \frac{(x+5)(x-2)}{(x+4)(x-4)}[/tex]
Now, putting the value of x = 2 in the expression,
⇒ (2+5)(2-2) / (2+4)(2-4)
⇒ 7*0 / 6*(-2)
= 0
Therefore, the limit exists.
For the other expressions, we cannot factorize the numerator or the denominator to cancel out the denominator because of which if we put the values of x directly into the denominator without cancelling it out, the denominator would be zero, which means infinity because zero in the denominator of a fraction means infinity or undefined or does not exist.
Hence, we can conclude that expressions 1 and 6 are limits that exist.
Learn more about limits here: https://brainly.com/question/13100668
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