Question

Can anyone solve this

Answer 1

Answer:

We have to compute the following limit.

$\displaystyle{\lim_{x→2} \: \: \frac{x - 2}{ \sqrt{ {x}^{2} - 4 } } }$

• Substitute x=2

$\displaystyle{\lim_{x→2} \: \: \frac{2 - 2}{ \sqrt{ {2}^{2} - 4 } } }$

$\displaystyle{\lim_{x→2} \: \: \frac{0}{ \sqrt{4 - 4} } }$

$\displaystyle{\lim_{x→2} \: \: \frac{0}{ \sqrt{0} } \: = \frac{0}{0} = 0}$

Using direct substitution to find the limit results in the indeterminate form $\displaystyle\frac{0}{0}$In order to evaluate the limit, we need to transform the expression to remove the indeterminate form. This is accomplished by using the relationship for the difference of squares of real numbers:

As for rational function with square root in it, we conjugate the expression by multiplying both denominator and numerator with the square root expression.

$\displaystyle{\lim_{x→2} \: \: \frac{(x - 2)}{ (\sqrt{ {x}^{2} - 4) } } \times \frac{( \sqrt{ {x}^{2} - 4 )} }{ (\sqrt{ {x}^{2} - 4) } } }$

$\displaystyle{\lim_{x→2} \: \: \frac{(x - 2)( \sqrt{ {x}^{2} - 4) } }{ \sqrt{ {x}^{2} - 4 } } }$

$\displaystyle{\lim_{x→2} \: \: \frac{ \cancel{(x - 2)}( \sqrt{ {x}^{2} + 4)} }{(x + 2) \cancel{(x - 2)}} }$

$\displaystyle{\lim_{x→2} \: \: \frac{ \sqrt{ {x}^{2} - 4 } }{x + 2)} }$

• Substitute x=2

$\displaystyle{ \frac{ \sqrt{ {2}^{2} - 4} }{2 + 2} = \frac{ \sqrt{4 - 4} }{4} = \frac{ \sqrt{0} }{4} = \frac{0}{4} = 0 }$

Thus,the limit value of expression is 0.

• Hint: In the above given type of questions first we will have to decide what limits are we going to apply and also check what is the function in the square root, there are different methods of solving limit of square root, like taking common of the highest root from the denominator as well as numerator, by rationalizing either denominator or numerator, by using L’Hospital’s Rule.

Answer 2

Answer:

0

Step-by-step explanation:

Hi! We are given the limit expression:

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

If we directly substitute x = 2 then we get 0/0 which is an indeterminate form. Therefore, we need to find other methods to evaluate the limit that does not become an indeterminate form.

As for rational function with square root in it, we conjugate the expression by multiplying both denominator and numerator with the square root expression.

$\displaystyle \large{ \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2-4})}{\sqrt{x^2-4}\sqrt{x^2-4}} }\\\displaystyle \large{ \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2-4})}{x^2-4} }$

When two same square root expressions multiply each other, the square root is taken out as shown above.

From denominator, we can factor x²-4 to (x-2)(x+2) via differences of two squares.

Hence:

$\displaystyle \large{ \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2-4})}{(x+2)(x-2)} }$

Cancel x-2.

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

Then substitute x = 2 which we receive 0/4 = 0.

Henceforth, the limit value of expression is 0.