MCQ
$\lim_\limits{\text{x} \rightarrow \text{a}}​\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}​(n\ \epsilon\ N)$ equals:
  • A
    $ \infty$
  • B
    $ -\infty$
  • C
    $0$
  • does not exist

Answer

Correct option: D.
does not exist
Left hand limit is$\lim_\limits{\text{x} \rightarrow \text{a}}​\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}​​​​​=-\infty$
And Right hand limit is $\lim_\limits{\text{x} \rightarrow \text{a}}​\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}​​​​​=+\infty$
$\text{L.H.L.}\neq \text{R.H.L.}$
Therefore, the given limit does not exist.
Hence, the option $D$ is correct.

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