f(x)=x^2-4/x-2,then show that limit exist
Answers
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
In this case we write
TO FIND
CALCULATION
RIGHT HAND LIMIT
LEFT HAND LIMIT
So
RESULT
Hence
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