Question

$\displaystyle \sf { solve}$

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


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Answer 1

$\large\text{\underline{Answer}}$

$\large 4$

$\large\text{\underline{Explanation}}$

If we directly substitute the value, it is in the form of $\dfrac{0}{0}$.

$\boxed{\text{ \dfrac{x^2-4}{x-2} at x=2 does not exist.}}$

By use of the properties of limits, we can eliminate the common factor existing in the fraction.

$\boxed{\begin{aligned}\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}&=\displaystyle\lim_{x\to2}\dfrac{(x+2)\cancel{(x-2)}}{\cancel{x-2}}\\\\&=\displaystyle\lim_{x\to2}(x+2)\ [\because\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)]\\\\&=4\end{aligned}}$

$\large\text{\underline{Extra explanation}}$

If the given function approaches $x=a$ from the left, the value it approaches is called the LHL(left-hand limit).

$\cdots\longrightarrow\displaystyle\text{ \lim_{x\to a-}f(x) is the LHL.}$

In the same manner, if the given function approaches from the right, the value it approaches is called the RHL(right-hand limit).

$\cdots\longrightarrow\displaystyle\text{ \lim_{x\to a+}f(x) is the RHL.}$

When $\text{LHL = RHL}$, the limit exists.

$\cdots\longrightarrow\displaystyle\text{ \lim_{x\to a}f(x) is the limit.}$

When $\text{LHL \neq RHL}$, the limit does not exist.

$\cdots\longrightarrow\displaystyle\text{Limit does not exist if \lim_{x\to a-}f(x)\neq\lim_{x\to a+}f(x) .}$

Answer 2

Answer:

Multiplynumeratoranddenominatorby(

x+2

+

3x−2

)

wewillget

⇒lt

x→2

(

x+2

−

3x−2

)×(

x+2

+

3x−2

)

(x

2

−4)(

x+2

+

3x−2

)

⇒lt

x→2

x+2−(3x−2)

(x

2

−4)(

x+2

+

3x−2

)

⇒lt

x→2

−2x+4

(x−2)(x+2)(

x+2

+

3x−2

)

⇒lt

x→2

−2

(x+2)(

x+2

+

3x−2

)

⇒

−2

(2+2)(

2+2

+

6−2

)

=

−2

4×4

=−8