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Estimate the limits numerically. \(\lim _{x \rightarrow 0} \frac{x^{2}}{x+1}\)

Short Answer

Expert verified
The limit as x approaches 0 of the function \(\frac{x^2}{x+1}\) is 0, which we can verify analytically and confirm numerically as the function approaches 0 for values very close to 0: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = 0 \]
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Step 1: Identify the limit to be evaluated

We are given the function \(\frac{x^2}{x+1}\), and we need to find the limit as x approaches 0. The limit we need to evaluate is: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} \]

Step 2: Substitute \(x\) with \(0\)

To check if this limit exists, substitute \(x\) with 0 in the function: \[ \frac{0^2}{0+1} = \frac{0}{1} = 0 \] Since the function evaluates to 0, the limit exists.

Step 3: Calculate the limit analytically

Now that we know the limit exists, we can calculate it analytically. In this case, the function is \(\frac{x^2}{x+1}\), and as x approaches 0, this simplifies to: \[ \frac{x^2}{x+1} \approx \frac{0^2}{0+1} \] \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = \frac{0}{1} = 0 \]

Step 4: Confirm the limit numerically

Finally, we can confirm the limit by substituting values very close to \(0\) and observing if the function continues to approach 0: For \(x = 0.1\): \[ \frac{0.1^2}{0.1+1} = \frac{0.01}{1.1} = 0.00909 \] For \(x = -0.1\): \[ \frac{-0.1^2}{-0.1+1} = \frac{0.01}{0.9} = 0.01111 \] As we can see, as the values of \(x\) get closer to \(0\), the function approaches \(0\). This confirms that the limit is indeed 0: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = 0 \]

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