find-the-limits-in-exercises-1-12-lim-_x-y-rightarrow0-ln-2-ex-y-3

Question

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

Stephen Hobbs · Accepted Answer

evaluating this limit for X approaches. Zero. Why approaches Elena too? We could use direct substitution, so we hav e to the zero or x minus Ln of to And then another way to rewrite this is we could think of it as e to the the zero goes away. And we can move whatever constant is in front of the natural log to the power. And so we could really imagine it as Ln of two to the power negative one. And then from here, we know that the, um, exponential and natural logger and verses of one another. And so our final solution would be to the negative one or one.

Stephen Hobbs · Answer

and this problem will start off with evaluating e to the X and using direct substitution will get eat 20 multiplied by sine of X, divided by acts. And in this case, we have before we even substitute. We could imagine as X is approaching zero here. There&#x27;s a rule that tells us that this especially if we use Low Patel it would be co sign of X over one and then substituting Zero would just get us one. So really, we could use an identity here. You to the zero would also be one which would evaluate this whole limit, so one.

Stephen Hobbs · Answer

using direct substitution for X is one. And why is one we would end up with the natural log of the absolute value of one plus one. Four X squared would just get its one and then multiply by. Y squared is also just one there. And so really, we&#x27;re going to end up with Ellen of the absolute value of two, and that&#x27;s our final solution.

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.

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 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.

Problem

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

Problem 7 Easy Difficulty

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

Answer

$\frac{1}{2}$

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$\frac{1}{2}$

Thomas Calculus

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Vectors Intro

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

Lily A.

Catherine R.

Samuel H.

Joseph L.

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

Thomas Calculus