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Practice with Limits

Here are some problems to practice what you have learned! If you need a hint on any of them, there's a few for each problem.

$\displaystyle{\lim_{x\to 0} x+1}$

Hint 1

Remember that our ultimate goal is to substitute $x$ for $0$ when we can. Is there anything preventing us from doing so?

Solution 1

We can plug in $x=0$ to the function, and see that

\lim_{x\to 0} x+1=0+1=1.

$\displaystyle{\lim_{x \to 2} \frac{x^2 -4}{x-2}}$

Hint 1

Always, when computing a limit, ask yourself: "Is there anything preventing me from just plugging in the value to the function?"

Hint 2

When you find what's preventing you from computing $\frac{x^2 -4}{x-2}$ when $x=2$ , try to eliminate it.

Solution 1

We can't compute $\frac{x^2 -4}{x-2}$ when $x=2$ because then we would be dividing by $0$ . However, we see that

f(x)=\frac{x^2 -4}{x-2}= \frac{(x -2)(x+2)}{x-2}.

Remember that we are not finding $f(2)$ , we are finding the limit of f(x) as x approaches 2. So, we can divide the top and bottom of the fraction by $(x−2)$ because we know $(x−2)$ is not equal to $0$ , but rather very, very, close. Hence, our answer is $x+2=2+2=4.$

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