Math, asked by tanu2321, 7 months ago

f(x)=x^2-4/x-2,then show that limit exist​

Answers

Answered by pulakmath007
18

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FORMULA TO BE IMPLEMENTED

CONDITION FOR EXISTENCE OF LIMIT

A function f(x) is said to have a limit L at a point x = a if

\displaystyle \lim_{x \to a + } \: f(x) =\displaystyle \lim_{x \to a - } f(x) = L

In this case we write

\displaystyle \lim_{x \to a} f(x) =  L

TO FIND

\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}

CALCULATION

RIGHT HAND LIMIT

\displaystyle \lim_{x \to 2 + } \frac{x^2-4}{x-2}

= \displaystyle \lim_{x \to 2 + } \frac{(x + 2)(x - 2)}{x-2}

= \displaystyle \lim_{x \to 2 + } {(x + 2)}

 = 2 + 2

 = 4

LEFT HAND LIMIT

\displaystyle \lim_{x \to 2  -  } \frac{x^2-4}{x-2}

= \displaystyle \lim_{x \to 2 + } \frac{(x + 2)(x - 2)}{x-2}

= \displaystyle \lim_{x \to 2  -  } {(x + 2)}

 = 2 + 2

 = 4

So

\displaystyle \lim_{x \to 2 + } \frac{x^2-4}{x-2} = \displaystyle \lim_{x \to 2 - } \frac{x^2-4}{x-2} = 4

RESULT

Hence

\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2} = 4

Answered by jiya91729
0

Answer:

FORMULA TO BE IMPLEMENTED

CONDITION FOR EXISTENCE OF LIMIT

A function f(x) is said to have a limit L ata point x = a if

\displaystyle \lim_{x \to a + } \: f(x) =\displaystyle \lim_{x \to a - } f(x) = Lx→a+limf(x)=x→a−limf(x)=L

In this case we write

\displaystyle \lim_{x \to a} f(x) = Lx→alimf(x)=L

TO FIND

\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2}x→2limx−2x2−4

CALCULATION

RIGHT HAND LIMIT

\displaystyle \lim_{x \to 2 + } \frac{x^2-4}{x-2}x→2+limx−2x2−4

= \displaystyle \lim_{x \to 2 + } \frac{(x + 2)(x - 2)}{x-2}=x→2+limx−2(x+2)(x−2)

= \displaystyle \lim_{x \to 2 + } {(x + 2)}=x→2+lim(x+2)

= 2 + 2=2+2

= 4=4

LEFT HAND LIMIT

\displaystyle \lim_{x \to 2 - } \frac{x^2-4}{x-2}x→2−limx−2x2−4

= \displaystyle \lim_{x \to 2 + } \frac{(x + 2)(x - 2)}{x-2}=x→2+limx−2(x+2)(x−2)

= \displaystyle \lim_{x \to 2 - } {(x + 2)}=x→2−lim(x+2)

= 2 + 2=2+2

= 4=4

So

\displaystyle \lim_{x \to 2 + } \frac{x^2-4}{x-2} = \displaystyle \lim_{x \to 2 - } \frac{x^2-4}{x-2} = 4x→2+limx−2x2−4=x→2−limx−2x2−4=4

RESULT

Hence

\displaystyle \lim_{x \to 2} \frac{x^2-4}{x-2} = 4x→2limx−2x2−4=4

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