Lim x- 2 - 1 x-2 - 4 x 3 - 2x 2 -

Now,We have, [ lim #160; #160;_ x→ 2 1 x 2 4 x^ 3 2 x^ 2 ...

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Standard XII

Mathematics

Algebra of Limits

lim x→ 2 [ 1 ...

Question

${lim}_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{3} - {2 x}^{2}}]$

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Solution

Now,We have,
$\begin{matrix} {lim}_{x \to 2} \frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}} = {lim}_{x \to 2} \frac{1}{x - 2} - \frac{4}{x^{2} (x - 2)} = {lim}_{x \to 2} \frac{x^{2} - 4}{x^{2} (x - 2)} = {lim}_{x \to 2} \frac{(x - 2) (x + 2)}{x^{2} (x - 2)} = {lim}_{x \to 2} \frac{x + 2}{x^{2}} = \frac{4}{4} = 1 \end{matrix}$

Hence, this is the answer.

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lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

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lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

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i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

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lim x \to 1 [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}]

${lim}_{x \to 2} (\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}})$

lim x \to \infty \frac{x^{3} + x^{2} + 1}{2 x^{2} + 3 x + 4}

$(x + 1)$ is a factor of the polynomial
(a) $x^{3} - 2 x^{2} + x + 2$ (b) $x^{3} + 2 x^{2} + x - 2$
(c) $x^{3} + 2 x^{2} - x - 2$ (d) $x^{3} + 2 x^{2} - x + 2$

Q. Solve

lim x \to 2 f (x) = lim x \to 2 \frac{(x^{2} - 4)}{(x - 2)}

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limx→2[1x−2−4x3−2x2]

Now,We have,limx→21x−2−4x3−2x2=limx→21x−2−4x2(x−2)=limx→2x2−4x2(x−2)=limx→2(x−2)(x+2)x2(x−2)=limx→2x+2x2=44=1Hence, this is the answer.

${lim}_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{3} - {2 x}^{2}}]$