Question

finding analytically the limit of a function f(x) at x=a and hence verifying the continuity of the function at that point.


eg: consider the function f(x) = x^2-4÷x-2, for x not equal to 2
4,. for x= 2.



PLEASE GIVE CORRECT ANSWER.​

Answer 1

Answer:

i hope this helps you ...

Answer 2

Tip:

If the numerator is the difference of two squares, and as such we can factorize using it as $A^2-B^2=(A-B)(A+B)$

Step

Step 1 of 2:

If we look at the graph of $y=\frac{x^2-4}{x-2}$  , we can see that it is clear that the limit exists, and is approximately $4$

 So, from the tip we can factorize as follows:

$=\lim_{x\rightarrow 2}\frac{x^2-4}{x-2}=\lim_{x\rightarrow 2}\frac{x^2-2^2}{x-2}\\\\=lim_{x\rightarrow 2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\rightarrow 2}(x+2)$

$=2+2\\\\=4$

Step 2 of 2:

Now for continuity:

L.H.L:

$=\lim_{x\rightarrow 2^-}\frac{x^2-4}{x-2}=\lim_{x\rightarrow 2^-}\frac{x^2-2^2}{x-2}\\\\=lim_{x\rightarrow 2^-}\frac{(x-2)(x+2)}{x-2}=\lim_{x\rightarrow 2^-}(x+2)$

$=\lim_{h\rightarrow 0}2-h+2\\\\=4$

R.H.L:

$=\lim_{x\rightarrow 2^+}\frac{x^2-4}{x-2}=\lim_{x\rightarrow 2^+}\frac{x^2-2^2}{x-2}\\\\=lim_{x\rightarrow 2^+}\frac{(x-2)(x+2)}{x-2}=\lim_{x\rightarrow 2^+}(x+2)$

$=\lim_{h\rightarrow 0}2+h+2\\\\=4$

L.H.L=R.H.L

S, the function is continuous .

Final Answer:

Function is continuous and have limit $4$.