L e t f ( x ) = { x 2 k ( x 2 − 4 ) 2 − x when x is an int eger otherwise then lim x → 2 f ( x ) Options: A . &nbsp exists only when k=1 B . &nbsp exists for every real k C . &nbsp exists for every real k = 1 D . &nbsp does not exist

Answer: Option B : B lim x → 2 + f ( x ) = lim x → 2 + k ( x 2 − 4 ) 2 − x = lim x → 2 + k ( − 2 − x ) = − 4 k lim x → 2 + f ( x ) = lim x → 2 + k ( x 2 − 4 ) 2 − x = lim x → 2 − k ( − 2 − x ) = − 4 k Hence limits exists for every real value of k.

Let f(x)={x2k(x2−4)2−xwhen x is an int eger otherwise then&

Discussion Forum : Limits Continuity And Differentiability

Question -

$L e t f (x) = {\begin{matrix} x^{2} & _{\frac{k (x^{2} - 4)}{2 - x}} \end{matrix} when x is an int eger otherwise then {lim}_{x \to 2} f (x)$

Options:

A . exists only when k=1

B . exists for every real k

C . exists for every real k = 1

D . does not exist

Answer: Option B
:
B

${lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{+}} k (- 2 - x) = - 4 k {lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{-}} k (- 2 - x) = - 4 k$

Hence limits exists for every real value of k.

Was this answer helpful ?

Next Question

Discussion Forum : Limits Continuity And Differentiability

Submit Your Solution hear: