How to compute a limit with exponential functions?
So i have this double variable limit
$$lim_{(x,y)→(0,0)} frac{e^{x+y}}{e^x-1}$$
I am new studying limits and my first thought was to plug in the x and y values, however it would get me an undefined solution , or else 1/0
Can somebody please show me the steps in order to compute the limit of this equation ? Thanks!
Edit: for reference, the solution on the answer sheet is as follows, but it doesn't really make any sense to me
The numerator clearly tends to 1, being a continuous function. The denumerator depends only on x and
goes to 0± as x goes to 0 from the right or from the left, that is, as (x, y) goes to (0, 0) staying in the left
half-plane or in the right half-plane (the function is not defined on the y axis). Therefore the limit is +∞ if
(x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
limits
asked Dec 12 '18 at 15:12
BM97BM97
758
|
show 4 more comments
So i have this double variable limit
$$lim_{(x,y)→(0,0)} frac{e^{x+y}}{e^x-1}$$
I am new studying limits and my first thought was to plug in the x and y values, however it would get me an undefined solution , or else 1/0
Can somebody please show me the steps in order to compute the limit of this equation ? Thanks!
Edit: for reference, the solution on the answer sheet is as follows, but it doesn't really make any sense to me
The numerator clearly tends to 1, being a continuous function. The denumerator depends only on x and
goes to 0± as x goes to 0 from the right or from the left, that is, as (x, y) goes to (0, 0) staying in the left
half-plane or in the right half-plane (the function is not defined on the y axis). Therefore the limit is +∞ if
(x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
limits
asked Dec 12 '18 at 15:12
BM97BM97
758
1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42
|
show 4 more comments
So i have this double variable limit
$$lim_{(x,y)→(0,0)} frac{e^{x+y}}{e^x-1}$$
I am new studying limits and my first thought was to plug in the x and y values, however it would get me an undefined solution , or else 1/0
Can somebody please show me the steps in order to compute the limit of this equation ? Thanks!
Edit: for reference, the solution on the answer sheet is as follows, but it doesn't really make any sense to me
The numerator clearly tends to 1, being a continuous function. The denumerator depends only on x and
goes to 0± as x goes to 0 from the right or from the left, that is, as (x, y) goes to (0, 0) staying in the left
half-plane or in the right half-plane (the function is not defined on the y axis). Therefore the limit is +∞ if
(x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
limits
asked Dec 12 '18 at 15:12
BM97BM97
758
So i have this double variable limit
$$lim_{(x,y)→(0,0)} frac{e^{x+y}}{e^x-1}$$
I am new studying limits and my first thought was to plug in the x and y values, however it would get me an undefined solution , or else 1/0
Can somebody please show me the steps in order to compute the limit of this equation ? Thanks!
Edit: for reference, the solution on the answer sheet is as follows, but it doesn't really make any sense to me
The numerator clearly tends to 1, being a continuous function. The denumerator depends only on x and
goes to 0± as x goes to 0 from the right or from the left, that is, as (x, y) goes to (0, 0) staying in the left
half-plane or in the right half-plane (the function is not defined on the y axis). Therefore the limit is +∞ if
(x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
limits
limits
asked Dec 12 '18 at 15:12
BM97BM97
758
asked Dec 12 '18 at 15:12
BM97BM97
758
edited Dec 12 '18 at 15:49
BM97
asked Dec 12 '18 at 15:12
BM97BM97
758
asked Dec 12 '18 at 15:12
BM97BM97
758
asked Dec 12 '18 at 15:12
BM97BM97
758
758
1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42
|
show 4 more comments
1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42
1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42
|
show 4 more comments
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1/0 is not really considered as undefined, rather it is worth $pm infty$ depending on wether $x$ is going to 0 from positive or negative values. Have you ever seen such limit ? What is your definition of limit by the way (rigorous with $epsilon$ or just behavior that one could see on a graph) ?
– Gâteau-Gallois
Dec 12 '18 at 15:29
I am not really sure. The question justs ask to compute the limit, with no further definitin
– BM97
Dec 12 '18 at 15:35
"... find the limit" does not make sense: the limit simply does not exist as you have observed. More or less the same as $lim_{xto 0}frac{1}{e^x-1}$ does not exist. What is your question?
– user587192
Dec 12 '18 at 15:39
Ok but you've never seen the definition of limit in some other context of your course ?
– Gâteau-Gallois
Dec 12 '18 at 15:39
My question reguards also how the solution explains the problem , orelse : Therefore the limit is +∞ if (x, y) → 0 and x > 0 and it is −∞ if (x, y) → 0 with x < 0.
– BM97
Dec 12 '18 at 15:42