Question

please refer to the attachment​

Answer 1

Answer:

sorry dono know the answer sorry

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 $f(x)=\frac{x^2-4}{x-2}$, we can see that it is clear that the limit exists, and is approximately $4$.

So, from the tip we can factorize as follows:

$=lim_{x\rightarrow2}{\frac{x^2-4}{x-2}}=\lim_{x\rightarrow2}{\frac{x^2-2^2}{x-2}}\\\\=\lim_{x\rightarrow2}\frac{(x-2)(x+2)}{x-2}\\\\=\lim_{x\rightarrow2}(x+2)\\\\=2+2\\\\=4$

Step 2 of 2:

Now for continuity:

L.H.L:

$=lim_{x\rightarrow2^-}{\frac{x^2-4}{x-2}}=\lim_{x\rightarrow2^-}{\frac{x^2-2^2}{x-2}}\\\\=\lim_{x\rightarrow2^-}\frac{(x-2)(x+2)}{x-2}\\\\=\lim_{x\rightarrow2^-}(x+2)\\\\=\lim_{h\rightarrow0}2-h+2\\\\=4$

R.H.L.:

$=lim_{x\rightarrow2^+}{\frac{x^2-4}{x-2}}=\lim_{x\rightarrow2^+}{\frac{x^2-2^2}{x-2}}\\\\=\lim_{x\rightarrow2^+}\frac{(x-2)(x+2)}{x-2}\\\\=\lim_{x\rightarrow2^+}(x+2)\\\\=\lim_{h\rightarrow0}2+h+2\\\\=4$

L.H.L=R.H.L

So, the function is continuous .

Final Answer:

Function is continuous and have limit $4$.