Evaluating Limits - Limits Practice Problems
Problem: Evaluating Limits #1
Evaluate the limit:
\Large{\lim_{x\to 3}\frac{5x^2-3x}{3x^3+3}}
Step 1:
The first step is to always try solving the limit by direct substitution. So, we will substitute 3 into the function and see what we get.
\Large{\lim_{x\to 3}\frac{5x^2-3x}{3x^3+3} = \frac{5(3)^2 – 3(3)}{3(3)^3 + 3} = \frac{36}{84} = \frac{3}{7}}
This was a simple problem which involved evaluating limits by direct substitution!
Solution Complete!
Step 1:
The first step is to always try solving the limit by direct substitution. So, we will substitute 3 into the function and see what we get.
This was a simple problem which involved evaluating limits by direct substitution!
Solution Complete!
Problem: Evaluating Limits #2
Evaluate the limit:
\Large{\lim_{x\to 0}\frac{3x^2-2x}{x}}
Step 1:
First, we will try and substitute the limit into the function. So, substituting the 0 into the function, we have:
\large{\lim_{x \to 0}\frac{3x^2 – 2x}{x} = \frac{3(0)^2 – 2(0)}{0} = \frac{0}{0}}
So, we end up with \frac{0}{0} which is an indeterminate form.
Step 2:
Since we ended up with an indeterminate form, we need to rewrite the original function in another way, and then try to solve the limit again.
Notice how we can factor out the x from the numerator, as follows:
\large{\lim_{x\to 0}\frac{3x^2-2x}{x} = \lim_{x \to 0}(\frac{x(3x-2)}{x}) }
\large{=\lim_{x \to 0} (3x – 2)}
As you see, with some algebraic manipulation, we were able to rewrite the original function.
Step 3:
Now we can simply plug in the limit and solve, as follows:
\large{\lim_{x\to 0}3x – 2 = 3(0) – 2 = -2}
Solution Complete!
Problem: Evaluating Limits #3
Evaluate the limit:
\Large{\lim_{h\to -2}\frac{h^2+3h+2}{h+2}}
Step 1:
First, substituting the limit directly into the function, gives us:
\Large{\lim_{h\to -2}\frac{h^2+3h+2}{h+2} = \frac{(-2)^2 + 3(-2) + 2}{-2 + 2} = \frac{0}{0}}
Hence, we have an indeterminate form, which means we must rewrite the function.
Step 2:
Problem: Evaluating Limits #4
Evaluate the limit:
\Large{\lim_{x\to \frac{\pi}{2}}\frac{sin(x)}{cos(x)-1}}
Step 1:
Let’s attempt to solve the limit by direct substitution. We have:
Problem: Evaluating Limits #5
Evaluate the limit:
\Large{\lim_{k\to 1}\frac{k-\sqrt{k}}{k^2-1}}
Step 1:
First, let’s try substituting the limit into the function, as shown:
\Large{\lim_{k\to 1}\frac{k-\sqrt{k}}{k^2-1} = \frac{1 – \sqrt{1}}{(1)^2 – 1} = \frac{0}{0}}
So, we end up with an indeterminate form. Thus, we must rewrite the function in some way before plugging in the limit.
Step 2:
When we are rewriting functions that involve roots, we want to do what is known as rationalizing the denominator. We do this by multiplying the function with it’s conjugate, as follows:
Solution Locked! Click to View!
Problem: Evaluating Limits #6
Evaluate the limit:
\Large{\lim_{x\to -2}\frac{x^2-4}{x+2}}
Step 1:
If we substitute the limit into the function, we get:
\Large{\lim_{x\to -2}\frac{x^2-4}{x+2} = \frac{(-2)^2 \:- \:4}{(-2) \:+\: 2} = \frac{0}{0}}
Of course, this is an indeterminate form, so we must change the function in some way, before substituting the limit.
Step 2:
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