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Algebra of Limits

Let fx = x+...

Question

Let $f (x) = {x + λ, x < 1,$
$2 x - 3, x \geq 1,$ if $lim x \to 1 f (x)$ exists, then find value of $λ$ .

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Solution

$L H L = {lim}_{h \to 0} f (1 - h) = (1 - h) + λ = 1 + λ R H L {lim}_{h \to 0} f (1 + h) - 3 = - 1 f o r l i m i t t o e x i s t, L H L = R H L 1 + λ = - 1 ∴ λ = - 2$

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Similar questions

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) = {\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 a x, & 1 \leq x < 2 \end{matrix}

lim x \to 1 f (x)

exists, then the value of

a

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

lim x \to - 1 f (x)

Q. Assume that exists

{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}

holds for certain interval containing that point

x = 1,

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

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