Limits at Infinity Practice Problems
Problem: Limits at Infinity #1
Evaluate the limit:
\Large{\lim_{x\to \infty}\frac{2x^2+3x+1}{x^2-4}}
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
\large{\lim_{x \to \infty}\frac{2x^2+3x+1}{x^2-4} = \frac{2(\infty)^2+3(\infty)+1}{(\infty)^2-4} = \frac{\infty}{\infty}}
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is x^2. So, we have:
\large{\lim_{x \to \infty}\frac{2x^2+3x+1}{x^2-4} = \lim_{x\to \infty}(\frac{x^2(2+\frac{3}{x}+\frac{1}{x^2})}{x^2(1-\frac{4}{x^2})})}
\large{ = \lim_{x\to \infty}(\frac{2+\frac{3}{x}+\frac{1}{x^2}}{1-\frac{4}{x^2}})}
Step 3:
Now, using the limit laws, we can evaluate the limit for each term as shown:
\large{ = \lim_{x\to \infty}(\frac{2+\frac{3}{x}+\frac{1}{x^2}}{1-\frac{4}{x^2}}) = \frac{\lim_{x\to \infty}(2) \:+\: \lim_{x \to \infty}(\frac{3}{x}) \:+\: \lim_{x\to \infty}(\frac{1}{x^2})}{\lim_{x\to \infty}(1)\:-\:\lim_{x\to \infty}(\frac{4}{x^2})}}
Our knowledge of rational functions can tell us the limits of the second and third term in the numerator and the second term in the denominator. We know these limits to be 0.
Hence, we have:
\large{ = \frac{2 + 0 + 0}{1 – 0} = 2}
Thus, the final answer is:
\large{\lim_{x \to \infty}\frac{2x^2+3x+1}{x^2-4} = 2}
Solution Complete!
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is x^2. So, we have:
Step 3:
Now, using the limit laws, we can evaluate the limit for each term as shown:
Our knowledge of rational functions can tell us the limits of the second and third term in the numerator and the second term in the denominator. We know these limits to be 0.
Hence, we have:
Thus, the final answer is:
Solution Complete!
Problem: Limits at Infinity #2
Evaluate the limit:
\Large{\lim_{x\to -\infty}\frac{e^x+2x}{e^x-3x}}
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
\large{\lim_{x \to -\infty}\frac{e^x+2x}{e^x-3x} = \frac{e^{-\infty}+2(-\infty)}{e^{-\infty}-3(-\infty)} = \frac{\infty}{\infty}}
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
Problem: Limits at Infinity #3
Evaluate the limit:
\Large{\lim_{k\to \infty}\frac{2k^3-5k^2+3}{3k^3+4k+1}}
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
\large{\lim_{k \to \infty}\frac{2k^3-5k^2+3}{3k^3+4k+1} = \frac{2(\infty)^3-5(\infty)^2+3}{3(\infty)^3+4(\infty)+1} = \frac{\infty}{\infty}}
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is k^3. So, we have:
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is k^3. So, we have:
Problem: Limits at Infinity #4
Evaluate the limit:
\Large{\lim_{x\to -\infty}\frac{2x^2+3x+1}{x^2-4x+5}}
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
\large{\lim_{x \to -\infty}\frac{2x^2+3x+1}{x^2-4x+5}}
\large{ = \frac{2(-\infty)^2+3(-\infty)+1}{(-\infty)^2-4(-\infty)+5} = \frac{\infty}{\infty}}
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is x^2. So, we have:
Problem: Limits at Infinity #5
Evaluate the limit:
\Large{\lim_{x\to \infty}\frac{3x^3-2x^2+5}{x^2+4x-1}}
Step 1:
Although \infty is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:
\large{\lim_{x \to \infty}\frac{3x^3-2x^2+5}{x^2+4x-2}}
\large{ = \frac{3(\infty)^3-2(\infty)^2+5}{(\infty)^2+4(\infty)-1} = \frac{\infty}{\infty}}
Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.
Step 2:
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