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Which of the following satisfies the validity of the limit of the function $\lim _{x \rightarrow p} f(x)=\frac{x^2-p^2}{x-p}$ at $x=p?$

Option: 1
$\lim _{x \rightarrow p+} f(x)=\lim _{x \rightarrow p^{-}} f(x)=p$

Option: 2
$\lim _{x \rightarrow p+} f(x)=\lim _{x \rightarrow p^{-}} f(x)=0$

Option: 3
$\lim _{x \rightarrow p+} f(x)=\lim _{x \rightarrow p^{-}} f(x)=2p$

Option: 4
None of the above

Answers (1)

best_answer

Note that the function $f(x)=\frac{x^2-p^2}{x-p}$ is not defined at $x=p$ as the denominator becomes zero.

But the limit of the function $\lim _{x \rightarrow p} f(x)=\frac{x^2-p^2}{x-p}$ may still exist at $x=p$ only when all the following hold well.

$\lim _{x \rightarrow p^{+}} f(x)$ is defined.
$\lim _{x \rightarrow p^{-}} f(x)$ is defined.
$\lim _{x \rightarrow p^{+}} f(x)=\lim _{x \rightarrow p^{-}} f(x)$

The provided limit can be reduced in the following way from the fact that $x \rightarrow p \Rightarrow x \neq p$

$\lim _{x \rightarrow p+} f(x)$

$=x^{2}\, - p^{2}\, /x-p$

$= (x-p) (x+p) / (x-p)$

$= x+p$ $[ x \neq p]$

Now, the Left Hand Limit

$\lim _{x \rightarrow p-} f(x)$

$=\lim _{h \rightarrow o} f(p-h)$

$=\lim _{h \rightarrow o} f(p-h+p)$

$=2p$

Now, the Right Hand Limit

$\lim _{x \rightarrow p+} f(x)$

$=\lim _{h \rightarrow o} f(p+h)$

$=\lim _{h \rightarrow o} f(p+h+p)$

$=2p$

Thus, $\lim _{x \rightarrow p+} f(x)=\lim _{x \rightarrow p^{-}} f(x)=2p$

Posted by

SANGALDEEP SINGH

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