
What is the limit of \[\underset{(x,y)\to (0,0)}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\] .
Answer
506.1k+ views
Hint: We will solve this question by solving this question in two paths. First to solve this we will take the path of x tending to zero in the limit and see what the answer we get in that case is and keep that in mind. For the second case that is the second path of the limit we take the path that y tends to zero in the limit and we therefore check the answer we get in that case. Now that we have gone from both paths to get the limit of the expression both the paths should give us the same answer and if it doesn’t then the limit does not exist and if it is the same for both cases then the answer we get is the limit of the expression.
Complete step by step solution:
To solve this question we begin with
For the limit for this expression to exist the expression must approach the same value of limit regardless of which path we go through to reach \[(0,0)\].
Now let’s say we use the path of the x axis to reach \[(0,0)\] which means we will fix y as zero to get to it. We can show this as
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+0}\]
Therefore
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}}\]
Dividing
\[\underset{x\to 0}{\mathop{\lim }}\,1\]
\[=1\]
Now after this let’s say we use the path of the y axis to reach \[(0,0)\] which means we will fix x as zero to get to it. We can show this as
\[\underset{y\to 0}{\mathop{\lim }}\,\dfrac{{{0}^{2}}}{{{0}^{2}}+{{y}^{2}}}\]
Therefore
\[\underset{y\to 0}{\mathop{\lim }}\,\dfrac{0}{{{y}^{2}}}\]
\[\underset{y\to 0}{\mathop{\lim }}\,0\]
Which means the limit will be \[=0\]
Now we can see that the origin along these two paths of x and y axis for the limit leads to different limits
\[\underset{x\to 0,y=0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\underset{y\to 0,x=0}{\mathop{\ne \lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
Therefore, the limit does not exist
Note: In simple words limit can be explained to be the value of a function when the input variable reaches some value. A common mistake that can be made in this question is sometimes some students just go solve both the paths together leading to errors in solution. Always solve the paths differently if we have more than one part and both should get you the same answer too or else the limit does not exist.
Complete step by step solution:
To solve this question we begin with
For the limit for this expression to exist the expression must approach the same value of limit regardless of which path we go through to reach \[(0,0)\].
Now let’s say we use the path of the x axis to reach \[(0,0)\] which means we will fix y as zero to get to it. We can show this as
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+0}\]
Therefore
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}}\]
Dividing
\[\underset{x\to 0}{\mathop{\lim }}\,1\]
\[=1\]
Now after this let’s say we use the path of the y axis to reach \[(0,0)\] which means we will fix x as zero to get to it. We can show this as
\[\underset{y\to 0}{\mathop{\lim }}\,\dfrac{{{0}^{2}}}{{{0}^{2}}+{{y}^{2}}}\]
Therefore
\[\underset{y\to 0}{\mathop{\lim }}\,\dfrac{0}{{{y}^{2}}}\]
\[\underset{y\to 0}{\mathop{\lim }}\,0\]
Which means the limit will be \[=0\]
Now we can see that the origin along these two paths of x and y axis for the limit leads to different limits
\[\underset{x\to 0,y=0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\underset{y\to 0,x=0}{\mathop{\ne \lim }}\,\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
Therefore, the limit does not exist
Note: In simple words limit can be explained to be the value of a function when the input variable reaches some value. A common mistake that can be made in this question is sometimes some students just go solve both the paths together leading to errors in solution. Always solve the paths differently if we have more than one part and both should get you the same answer too or else the limit does not exist.
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