Let f x -1-2xa- 0- x- 1 ax- 1- x- 2If limx- 1f x exist then a is

The correct options are B 1 C 2 f(x)=1+ 2x a for 0 lt;x lt;1 =ax for 1≤ x lt;2 As the function is continuous ...

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Factorization Method Form to Remove Indeterminate Form

Let f x =1+...

Question

Let $f (x) = {\begin{matrix} 1 + \frac{2 x}{a}; 0 \leq x < 1 a x, 1 \leq x < 2 \end{matrix}$
If $lim x \to 1 f (x)$ exist then $a$ is

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Similar questions

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1 + \frac{2 x}{a}, & 0 \leq x < 1 a x, & 1 \leq x < 2 \end{matrix}

lim x \to 1 f (x)

a

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} 1 + \frac{2 x}{a}, 0 \leq x < 1 \frac{x}{a}, 1 \leq x \leq 2 \end{matrix} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

lim x \to 1 f (x)

f (x) = {x + λ, x < 1,

2 x - 3, x \geq 1,

lim x \to 1 f (x)

λ

f (x) = \begin{matrix} x + 2 & x \leq - 1 c x^{2} & x > - 1 \end{matrix}

lim x \to - 1 f (x)

{lim}_{x \to - 1} f (x)

\frac{x^{2} + x - 2}{x + 3} \leq \frac{f (x)}{x^{2}} \leq \frac{x^{2} + 2 x - 1}{x + 3}

x = 1,

{lim}_{x \to - 1} f (x)

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