find-the-limits-in-exercises-1-12-lim-_x-y-rightarrow00-frac3-x2-y25x2y22-3

Question

Find the limits in Exercises $1-12$.
$$
\lim _{(x, y) \rightarrow(0,0)} \frac{3 x^{2}-y^{2}+5}{x^{2}+y^{2}+2}
$$

Stephen Hobbs · Accepted Answer

we&#x27;re going to evaluate this limit. We could go ahead and use direct substitution substitution just like we would in a regular one variable limit problem. So we&#x27;ll go ahead and use X zero and Wyatt zero here. So this is going to be 30 squared minus zero squared, and then we have plus five up top. And so you might already be able to see everything&#x27;s going to get knocked out except that five in the numerator. And then in the denominator, it&#x27;s just going to be zero dsquared and mine plus zero squared. And again, the only thing in the denominator that it&#x27;s going to remain. That&#x27;s going to be that, too. And so therefore, we&#x27;re gonna end up with this limit having a value of 2/5.

Stephen Hobbs · Answer

in this multi variable limit problem, we could use direct substitution. Justus we would with a one variable limit problem. And so we&#x27;ll end up getting 1/2 plus a negative 1/3. Or we could just go ahead and say minus but 1/3. And then we might as well just go ahead and convert these so we could do it all by hand and we&#x27;ll come up with our common denominator here, multiplying top bottom the first fraction but three. So we&#x27;ll end up with 36 multiply second fraction, top bottom by two minus 26 And so, at this point, you could see we&#x27;ll end up getting 1/6 all squared. And from there that would just go ahead. And where the numerator spread the denominators or a final solution is 1/36.

Zachary Mitchell · Answer

we have the limit as X goes to two from the left hand side. Now, notice that the denominator here is tending to zero, and it&#x27;s tending to zero. Um, so that s so that the denominator is always negative because we&#x27;re approaching for the left hand side. The denominator is always negative as it approaches zero. So we have that this approach is negative Infinity.

Zachary Mitchell · Answer

we have the limit as X goes to zero from the left hand side of 5/2 X. That means that the denominator is becoming very small and as it&#x27;s becoming very small, it&#x27;s nay. It has a negative value, so this means that the fraction is tending to negative infinity.

Zachary Mitchell · Answer

we have the limits as X approaches zero of one over X to the to thirds. Now notice that as X approaches zero, the denominator here approaches zero and the denominator is always greater than or equal to zero because the exponents is two thirds. Therefore, this limit is equal to infinity.

Zachary Mitchell · Answer

we have the limit as X approaches zero of negative one. Divide by X squared X plus one. Now in the denominator, we have an X squared that X squared is approaching zero and it&#x27;s always greater than or equal to zero as X approaches zero and notice that X plus one is is approaching one as X approaches. Zero. So ultimately we have that the denominator is approaching zero and is always greater than or equal to zero as X approaches zero and the numerator is a negative number. So we have negative. Infinity is the limit.

Problem

Find the limits in Exercises $1-12$. $$ \lim _{…

Problem 1 Easy Difficulty

Find the limits in Exercises $1-12$.
$$
\lim _{(x, y) \rightarrow(0,0)} \frac{3 x^{2}-y^{2}+5}{x^{2}+y^{2}+2}
$$

Answer

$\frac{5}{2}$

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$\frac{5}{2}$

Thomas Calculus

Caleb E.

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Vectors Intro

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George B. Thomas, Maurice D. Weir, Joel Hass, Frank R. GiordanoThomas Calculus

Caleb E.

Kristen K.

Samuel H.

Michael J.

George B. Thomas, Maurice D. Weir, Joel Hass, Frank R. Giordano

Thomas Calculus