📚Limits Solved Problems
Problem 1Find the limit of the following expression using direct substitution: \[\lim_{(x,y)\to (0, 0)} \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to (0, 0)} \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1}\).
Substitute \(x=0\) and \(y=0\):
\begin{align*}\lim_{(x,y)\to (0, 0)} \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} &= 0\end{align*}
Step-1: Understand direct substitution.
For continuous functions, we can find the limit by simply substituting the point into the expression. This is the simplest method to check a limit.
Step-2: Substitute the limit point.
Replace \(x=0\) and \(y=0\) in the expression.
\begin{align*}&= 0\end{align*}
Step-3: Interpret the result.
The expression evaluates to a finite number. Since the function appears to be continuous at this point, the limit exists and equals this value.
Problem 2Find the limit of the following expression using direct substitution: \[\lim_{(x,y)\to (0, 0)} e^{- x^{2} - y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to (0, 0)} e^{- x^{2} - y^{2}}\).
Substitute \(x=0\) and \(y=0\):
\begin{align*}\lim_{(x,y)\to (0, 0)} e^{- x^{2} - y^{2}} &= 1\end{align*}
Step-1: Understand direct substitution.
For continuous functions, we can find the limit by simply substituting the point into the expression. This is the simplest method to check a limit.
Step-2: Substitute the limit point.
Replace \(x=0\) and \(y=0\) in the expression.
\begin{align*}&= 1\end{align*}
Step-3: Interpret the result.
The expression evaluates to a finite number. Since the function appears to be continuous at this point, the limit exists and equals this value.
Problem 3Find the following limit using algebraic simplification: \[\lim_{(x,y)\to (0,0)} \frac{x^{3} - y^{3}}{x - y}\] \textit{Hint: Use factorization, cancellation, or rationalization.
Quick Solution
Detailed Solution
Step-1: Apply algebraic simplification.
Apply algebraic simplification.
Using cancellation:
\begin{align*}\frac{x^{3} - y^{3}}{x - y} &= x^{2} + x y + y^{2}\end{align*}
Step-2: Evaluate the limit.
Evaluate the limit.
The limit can now be evaluated.
Step-1: Examine the original expression.
We are given: \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^{3} - y^{3}}{x - y}\)
If we try direct substitution, we get an indeterminate form. This tells us we need to simplify the expression first.
Step-2: Look for algebraic simplification techniques.
Common techniques for multivariable limits include:
Step-3: Apply the simplification technique.
We can cancel common factors between numerator and denominator.
This is valid because we're taking a limit, not evaluating at the point:
\begin{align*}\frac{x^{3} - y^{3}}{x - y} &= x^{2} + x y + y^{2}\end{align*}
Step-4: Verify the simplification.
The simplified expression is equivalent to the original for all \((x,y) \neq (0,0)\). We can check by multiplying back or by testing a few points.
Step-5: Evaluate the limit using the simplified expression.
After simplification, we can now evaluate the limit.
Problem 4Find the following limit using algebraic simplification: \[\lim_{(x,y)\to (0,0)} \frac{x^{2} + 2 x y + y^{2}}{x + y}\] \textit{Hint: Use factorization, cancellation, or rationalization.
Quick Solution
Detailed Solution
Step-1: Apply algebraic simplification.
Apply algebraic simplification.
Using factorization:
\begin{align*}\frac{x^{2} + 2 x y + y^{2}}{x + y} &= x + y\end{align*}
Step-2: Evaluate the limit.
Evaluate the limit.
The limit can now be evaluated.
Step-1: Examine the original expression.
We are given: \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^{2} + 2 x y + y^{2}}{x + y}\)
If we try direct substitution, we get an indeterminate form. This tells us we need to simplify the expression first.
Step-2: Look for algebraic simplification techniques.
Common techniques for multivariable limits include:
Step-3: Apply the simplification technique.
We notice that the numerator and denominator have common factors.
Let's factor both the numerator and denominator:
\begin{align*}\text{Denominator: } & \text{factors to } ...\end{align*}
After factoring, we can cancel the common factors:
\begin{align*}\frac{x^{2} + 2 x y + y^{2}}{x + y} &= x + y\end{align*}
Step-4: Verify the simplification.
The simplified expression is equivalent to the original for all \((x,y) \neq (0,0)\). We can check by multiplying back or by testing a few points.
Step-5: Evaluate the limit using the simplified expression.
After simplification, we can now evaluate the limit.
Problem 5Find the limit of the following expression using known limit patterns: \[\lim_{(x,y)\to (0,0)} \frac{\sin{\left(x^{2} + y^{2} \right)}}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{\sin{\left(x^{2} + y^{2} \right)}}{x^{2} + y^{2}}\).
Recognize the standard limit pattern:
\(\frac{\sin u}{u} \to 1\)
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} &= 1 \\\lim_{u\to0} \frac{e^u - 1}{u} &= 1 \\\lim_{u\to0} \frac{\ln(1+u)}{u} &= 1\end{align*}
Step-1: Identify the pattern.
The expression matches a known limit pattern:
Step-2: Introduce substitution.
Let
\begin{align*}u = x^{2} + y^{2}\end{align*}
As \((x,y) \to (0,0)\) we have \(u \to 0\).
Step-3: Rewrite the limit.
\begin{align*}&= 1\end{align*}
Step-4: Apply the known limit.
We know that
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} = 1\end{align*}
Step-5: State the final result.
Therefore the original limit equals:
Problem 6Find the limit of the following expression using known limit patterns: \[\lim_{(x,y)\to (0,0)} \frac{e^{x^{2} + y^{2}} - 1}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{e^{x^{2} + y^{2}} - 1}{x^{2} + y^{2}}\).
Recognize the standard limit pattern:
\(\frac{e^u - 1}{u} \to 1\)
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} &= 1 \\\lim_{u\to0} \frac{e^u - 1}{u} &= 1 \\\lim_{u\to0} \frac{\ln(1+u)}{u} &= 1\end{align*}
Step-1: Identify the pattern.
The expression matches a known limit pattern:
Step-2: Introduce substitution.
Let
\begin{align*}u = x^{2} + y^{2}\end{align*}
As \((x,y) \to (0,0)\) we have \(u \to 0\).
Step-3: Rewrite the limit.
\begin{align*}&= 1\end{align*}
Step-4: Apply the known limit.
We know that
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} = 1\end{align*}
Step-5: State the final result.
Therefore the original limit equals:
Problem 7Find the limit of the following expression using known limit patterns: \[\lim_{(x,y)\to (0,0)} \frac{\log{\left(x^{2} + y^{2} + 1 \right)}}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{\log{\left(x^{2} + y^{2} + 1 \right)}}{x^{2} + y^{2}}\).
Recognize the standard limit pattern:
\(\frac{\ln(1+u)}{u} \to 1\)
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} &= 1 \\\lim_{u\to0} \frac{e^u - 1}{u} &= 1 \\\lim_{u\to0} \frac{\ln(1+u)}{u} &= 1\end{align*}
Step-1: Identify the pattern.
The expression matches a known limit pattern:
Step-2: Introduce substitution.
Let
\begin{align*}u = x^{2} + y^{2}\end{align*}
As \((x,y) \to (0,0)\) we have \(u \to 0\).
Step-3: Rewrite the limit.
\begin{align*}&= 1\end{align*}
Step-4: Apply the known limit.
We know that
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} = 1\end{align*}
Step-5: State the final result.
Therefore the original limit equals:
Problem 8Find the limit of the following expression using known limit patterns: \[\lim_{(x,y)\to (0,0)} \frac{1 - \cos{\left(x^{2} + y^{2} \right)}}{\left(x^{2} + y^{2}\right)^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{1 - \cos{\left(x^{2} + y^{2} \right)}}{\left(x^{2} + y^{2}\right)^{2}}\).
Recognize the standard limit pattern:
\(\frac{1-\cos u}{u^2} \to \frac{1}{2}\)
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} &= 1 \\\lim_{u\to0} \frac{e^u - 1}{u} &= 1 \\\lim_{u\to0} \frac{\ln(1+u)}{u} &= 1\end{align*}
Step-1: Identify the pattern.
The expression matches a known limit pattern:
Step-2: Introduce substitution.
Let
\begin{align*}u = x^{2} + y^{2}\end{align*}
As \((x,y) \to (0,0)\) we have \(u \to 0\).
Step-3: Rewrite the limit.
\begin{align*}&= \frac{1}{2}\end{align*}
Step-4: Apply the known limit.
We know that
\begin{align*}\lim_{u\to0} \frac{\sin u}{u} = 1\end{align*}
Step-5: State the final result.
Therefore the original limit equals:
Problem 9Find the limit of the following expression using the path test: \[\lim_{(x,y)\to (0,0)} \frac{- x^{2} + y^{2}}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to (0,0)} \frac{- x^{2} + y^{2}}{x^{2} + y^{2}}\)
Path Test Results:
Along \(x = 0\) (approach along y-axis): \(\displaystyle\lim_{(x,y)\to(0,0)} 1 = 1\)
Along \(y = 0\) (approach along x-axis): \(\displaystyle\lim_{(x,y)\to(0,0)} -1 = -1\)
Along \(y = mx\) (approach along any line through origin): \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{m^{2} x^{2} - x^{2}}{m^{2} x^{2} + x^{2}} = \frac{m^{2} - 1}{m^{2} + 1}\)
Conclusion: The limit along \(y=mx\) depends on \(m\), giving different values for different slopes. Therefore, the limit **does not exist**.
Step-1: Understanding the Path Test Method.
For a limit to exist at a point, the function must approach the same value regardless of the path taken to approach that point. If we can find two different paths that give different limits, then the limit does not exist.
We will test three simple paths:
Step-2: Test path along the y-axis ($x = 0$).
Substitute \(x = 0\) into the original expression:
\begin{align*}&= 1\end{align*}
Step-3: Test path along the x-axis ($y = 0$).
Substitute \(y = 0\) into the original expression:
\begin{align*}&= -1\end{align*}
Step-4: Test path along any line through the origin ($y = mx$).
Substitute \(y = mx\) into the original expression, where \(m\) is any constant (the slope):
\begin{align*}&= \frac{m^{2} - 1}{m^{2} + 1}\end{align*}
Notice that this limit **depends on the value of \(m\)** (the slope). Different slopes give different limit values.
For example:
Step-5: Compare the results from different paths.
We can see that the paths give different limit values:
The limit along \(y=mx\) is \(\frac{m^{2} - 1}{m^{2} + 1}\), which depends on \(m\). Since different values of \(m\) (different directions) give different limits, the function approaches different values from different directions.
Since a limit must be the same regardless of the path of approach, and we have found paths that give different values, we can conclude that the limit **does not exist**.
Problem 10Find the limit of the following expression using the path test: \[\lim_{(x,y)\to (0,0)} \frac{x^{2} y}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to (0,0)} \frac{x^{2} y}{x^{2} + y^{2}}\)
Path Test Results:
Along \(x = 0\) (approach along y-axis): \(\displaystyle\lim_{(x,y)\to(0,0)} 0 = 0\)
Along \(y = 0\) (approach along x-axis): \(\displaystyle\lim_{(x,y)\to(0,0)} 0 = 0\)
Along \(y = mx\) (approach along any line through origin): \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{m x^{3}}{m^{2} x^{2} + x^{2}} = 0\)
Conclusion: All paths approach the same value \(0\), so the limit may exist.
Step-1: Understanding the Path Test Method.
For a limit to exist at a point, the function must approach the same value regardless of the path taken to approach that point. If we can find two different paths that give different limits, then the limit does not exist.
We will test three simple paths:
Step-2: Test path along the y-axis ($x = 0$).
Substitute \(x = 0\) into the original expression:
\begin{align*}&= 0\end{align*}
Step-3: Test path along the x-axis ($y = 0$).
Substitute \(y = 0\) into the original expression:
\begin{align*}&= 0\end{align*}
Step-4: Test path along any line through the origin ($y = mx$).
Substitute \(y = mx\) into the original expression, where \(m\) is any constant (the slope):
\begin{align*}&= 0\end{align*}
Step-5: Compare the results from different paths.
All three paths give the same value \(0\). This suggests that the limit may exist, but we need to be cautious - there could still be some other path that gives a different value. Further testing with polar coordinates would be needed for confirmation.
Problem 11Find the limit of the following expression: \[\lim_{(x,y)\to (0,0)} \frac{- x^{2} + y^{2}}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Step-1: Convert to polar coordinates
Convert to polar coordinates
Let \(x = r\cos\theta\), \(y = r\sin\theta\). Then:
Step-2: Take limit as $r \to 0$
Take limit as \(r \to 0\)
The expression depends on \(\theta\), so the limit depends on the direction of approach.
Step-1: Understanding polar coordinates.
Polar coordinates are useful for limits because they allow us to approach the origin from all directions simultaneously. The transformation is:
\begin{align*}y &= r\sin\theta\end{align*}
Here, \(r\) represents the distance from the origin, and \(\theta\) represents the direction (angle). As \((x,y) \to (0,0)\), we have \(r \to 0\), while \(\theta\) can be any angle.
Step-2: Substitute polar coordinates into the expression.
\begin{align*}&= \lim_{r\to 0} - \cos{\left(2 \theta \right)}\end{align*}
Step-4: Check if the expression depends on $\theta$.
The simplified expression - \cos{\left(2 \theta \right)} contains \(\theta\). This means that as \(r \to 0\), the value of the expression depends on the angle \(\theta\), i.e., it depends on the direction from which we approach the origin.
For example:
Since the limit would depend on the direction of approach, it cannot exist.
Step-5: Take the limit as $r \to 0$.
Because the expression depends on \(\theta\), the limit as \(r \to 0\) is not well-defined (it would depend on which \(\theta\) we choose).
Step-6: Final conclusion.
The limit depends on the direction of approach (it depends on \(\theta\)). Therefore, the limit does not exist.
Problem 12Find the limit of the following expression: \[\lim_{(x,y)\to (0,0)} \frac{x^{2} y}{x^{2} + y^{2}}\]
Quick Solution
Detailed Solution
Step-1: Convert to polar coordinates
Convert to polar coordinates
Let \(x = r\cos\theta\), \(y = r\sin\theta\). Then:
Step-2: Take limit as $r \to 0$
Take limit as \(r \to 0\)
As \(r \to 0\), the expression approaches \(0\), independent of \(\theta\).
Step-1: Understanding polar coordinates.
Polar coordinates are useful for limits because they allow us to approach the origin from all directions simultaneously. The transformation is:
\begin{align*}y &= r\sin\theta\end{align*}
Here, \(r\) represents the distance from the origin, and \(\theta\) represents the direction (angle). As \((x,y) \to (0,0)\), we have \(r \to 0\), while \(\theta\) can be any angle.
Step-2: Substitute polar coordinates into the expression.
\begin{align*}&= \lim_{r\to 0} r \sin{\left(\theta \right)} \cos^{2}{\left(\theta \right)}\end{align*}
Step-4: Check if the expression depends on $\theta$.
The simplified expression r \sin{\left(\theta \right)} \cos^{2}{\left(\theta \right)} does not contain \(\theta\). This means that as \(r \to 0\), the expression approaches the same value regardless of the direction (angle) of approach.
Step-5: Take the limit as $r \to 0$.
\begin{align*}\lim_{r\to 0} r \sin{\left(\theta \right)} \cos^{2}{\left(\theta \right)} = 0\end{align*}
This value is independent of \(\theta\), meaning it's the same from all directions.
Step-6: Final conclusion.
Since the limit as \(r \to 0\) exists and is independent of \(\theta\), the original limit exists and equals \(0\).
Problem 13Find the limit of the following expression using the Squeeze Theorem: \[\lim_{(x,y)\to (0,0)} \left(x^{2} + y^{2}\right) \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)}\]
Quick Solution
Detailed Solution
Solution.
Given \(\displaystyle\lim_{(x,y)\to(0,0)} \left(x^{2} + y^{2}\right) \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)}\).
Using the Squeeze Theorem.
We use the inequality
\begin{align*}-1 \leq \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)} \leq 1\end{align*}
Multiplying by the factor \(x^{2} + y^{2}\) gives
\begin{align*}- \left|{x^{2} + y^{2}}\right| \leq \left(x^{2} + y^{2}\right) \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)} \leq \left|{x^{2} + y^{2}}\right|\end{align*}
Now compute the limit of the bounding functions:
\begin{align*}\lim_{(x,y)\to(0,0)} x^{2} + y^{2} = 0\end{align*}
Since both bounds approach \(0\), by the Squeeze Theorem we obtain
\begin{align*}\lim_{(x,y)\to(0,0)} \left(x^{2} + y^{2}\right) \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)} = 0\end{align*}
Step-1: Understanding the Squeeze Theorem.
The Squeeze Theorem states that if we can find two functions \(g(x,y)\) and \(h(x,y)\) such that \(g(x,y) \leq f(x,y) \leq h(x,y)\) near the point and \(\lim g(x,y) = \lim h(x,y) = L\), then \(\lim f(x,y) = L\).
This is particularly useful for functions involving bounded oscillations multiplied by a factor that goes to zero.
Step-2: Identify the bounded trigonometric part.
The expression contains \(\sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)}\) which is bounded:
\begin{align*}-1 \leq \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)} \leq 1\end{align*}
This is a key property of sine and cosine functions.
Step-3: Multiply the inequality by the factor.
The factor \(x^{2} + y^{2}\) multiplies the trigonometric part. When we multiply the inequality, we must be careful about the sign of the multiplier. In this case, we get:
\begin{align*}- \left|{x^{2} + y^{2}}\right| \leq \left(x^{2} + y^{2}\right) \sin{\left(\frac{1}{x^{2} + y^{2} + 1} \right)} \leq \left|{x^{2} + y^{2}}\right|\end{align*}
Step-4: Analyze the bounding functions.
Now we need to find the limits of the lower bound and upper bound as \((x,y) \to (0,0)\):
\begin{align*}\lim_{(x,y)\to(0,0)} \left|{x^{2} + y^{2}}\right| &= \lim_{(x,y)\to(0,0)} x^{2} + y^{2} \cdot (1)\end{align*}
Both of these limits depend on the limit of \(x^{2} + y^{2}\).
Step-5: Compute the limit of the multiplier.
\begin{align*}\lim_{(x,y)\to(0,0)} x^{2} + y^{2} = 0\end{align*}
This can be shown by direct substitution or path testing.
Step-6: Apply the Squeeze Theorem.
Since both bounds approach \(0\), we have:
\begin{align*}\lim_{(x,y)\to(0,0)} - \left|{x^{2} + y^{2}}\right| = 0 \quad \text{and} \quad \lim_{(x,y)\to(0,0)} \left|{x^{2} + y^{2}}\right| = 0\end{align*}
By the Squeeze Theorem, the original l
