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Problem 1

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

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: $\lim_{x \rightarrow 0} \frac{x^2}{x+1} = 0$
See the step by step solution

## 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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