Solve limit (as x approaches 2) of frac{sqrt{x-1}-1}{sqrt{x+2}-2} | Microsoft Math Solver

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5 problems similar to:

x \to 2 lim \frac{x - 1 - 1}{x + 2 - 2}

Similar Problems from Web Search

x \to 0 lim \frac{x + 1 - 1}{3 x + 1 - 1}

https://socratic.org/questions/5a5b02637c01497b6f0c8e0b

x \to 0 lim \frac{x + 1 - 1}{3 x + 1 - 1} = \frac{3}{2}

Explanation: Let:

t = 6 x + 1

x \to 0 lim \frac{x + 1 - 1}{3 x + 1 - 1} = t \to 1 lim \frac{t ^{3} - 1}{t ^{2} - 1}

x \to 4 lim (\frac{2 x - 7 - 1}{x - 3 - 1})

https://socratic.org/questions/ex-with-lim-lim-x-4-sqrt-2x-7-1-sqrt-x-3-1

The limit equals

2

Explanation: We have using L'hospitals:

L = x \to 4 lim \frac{\frac{2}{2 2 x - 7}}{\frac{1}{2 x - 3}}

How would I evaluate ?

https://www.quora.com/How-would-I-evaluate-lim_-limits-x-rightarrow-2-frac-sqrt-27-x-5-sqrt-18-x-4

Since there's already an answer to this question using l'Hopital's rule, I've decided to take another approach not involving differentiation Let's first try to manipulate the term a bit to simplify ...

x \to 1 lim \frac{x + 3 - 2}{x + 8 - 3}

https://math.stackexchange.com/questions/319188/find-displaystyle-lim-x-to-1-frac-sqrtx3-2-sqrtx8-3

\frac{x + 3 - 2}{x + 8 - 3} = (\frac{x + 3 - 2}{x + 8 - 3} \frac{x + 3 + 2}{x + 8 + 3}) \frac{x + 8 + 3}{x + 3 + 2}

Limit of quotients with square roots:

lim_{x \to 2} \frac{6 - x - 2}{3 - x - 1}

https://math.stackexchange.com/questions/195532/limit-of-quotients-with-square-roots-lim-x-to2-frac-sqrt6-x-2-sqrt3

You are right to try conjugates:

lim_{x \to 2} \frac{6 - x - 2}{3 - x - 1} = lim_{x \to 2} \frac{6 - x - 2}{3 - x - 1} \frac{3 - x + 1}{3 - x + 1} = lim_{x \to 2} \frac{( 6 - x ) ( 3 - x ) + 6 - x - 2 3 - x - 2}{2 - x}

Help me solve the limits

https://math.stackexchange.com/q/1601812

2)

lim_{x \to 0} \frac{\frac{( e ^{3 x} - 1 ) s i n 3 x}{3 x \cdot 3 x}}{\frac{l n ( 2 x ^{2} + 1 )}{9 x ^{2}}} = lim_{x \to 0} \frac{\frac{e ^{3 x - 1}}{3 x} \cdot \frac{s i n 3 x}{3 x}}{\frac{l n ( 2 x ^{2} + 1 )}{9 x ^{2}}} = lim_{x \to 0} \frac{1}{\frac{l n ( 2 x ^{2} + 1 )}{9 x ^{2}}} = lim_{x \to 0} \frac{1}{\frac{1}{9 x ^{2}} l n ( 2 x ^{2} + 1 )} = lim_{x \to 0} \frac{1}{l n ( 2 x ^{2} + 1 ) ^{\frac{1}{9 x ^{2}}}} = lim_{x \to 0} \frac{1}{l n ( 2 x ^{2} + 1 ) ^{\frac{1}{2 x ^{2}} \cdot \frac{2}{9}}} = \frac{1}{l n e ^{\frac{2}{9}}} = \frac{9}{2}

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