Find the following limits.
\lim\limits_{x \to \infty} \left(3x^3-2x^2+4\right)
The limit of a polynomial as x approaches infinity or negative infinity depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore, we have:
\lim\limits_{x \rightarrow \infty} \left(3x^3 - 2x^2 + 4\right) = \lim\limits_{x \rightarrow \infty} 3x^3 = 3\lim\limits_{x \rightarrow \infty} x^3 .
We know that:
\lim\limits_{x \rightarrow \infty} x^3 = \infty
Thus:
\lim\limits_{x \to \infty} f\left(x\right)=\infty
\lim\limits_{x \to \infty} \left(-x^2 + 2x -1\right)
The limit of a polynomial as x approaches infinity (or negative infinity) depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore we have:
\lim\limits_{x \rightarrow \infty} \left(-x^2 + 2x -1\right) = \lim\limits_{x \rightarrow \infty} -x^2 = -\lim\limits_{x \rightarrow \infty} x^2
We know that:
\lim\limits_{x \rightarrow \infty} x^2 = \infty
Therefore, remembering the negative sign of the coefficient:
\lim\limits_{x \to \infty} f\left(x\right)=-\infty
\lim\limits_{x \to 4} \left(2x^2 - 4x + 5\right)
To find the limit, plug 4 into the function:
\lim\limits_{x \rightarrow 4} f\left(x\right) = 2\left(4\right)^2 - 4\left(4\right) + 5 = 2\left(16\right) - 16 + 5 = 16 + 5 = 21
Therefore:
\lim\limits_{x \rightarrow 4} f\left(x\right) = 21
\lim\limits_{x \to \infty} f\left(x\right)=21
\lim\limits_{x \to \infty} \left(2x^2 - 4x + 5\right)
The limit of a polynomial as x approaches infinity (or negative infinity) depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore we have:
\lim\limits_{x \rightarrow \infty} \left(2x^2 - 4x + 5\right)= \lim\limits_{x \rightarrow \infty} 2x^2= 2\lim\limits_{x \rightarrow \infty} x^2
We know that:
\lim\limits_{x \rightarrow \infty} x^2 = \infty
Therefore:
\lim\limits_{x \to \infty} f\left(x\right)=\infty
\lim\limits_{x \to -\infty} \left(-5x^7 + x^4 + 3x^2 - x\right)
The limit of a polynomial as x approaches infinity (or negative infinity) depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore we have:
\lim\limits_{x \rightarrow -\infty} \left(-5x^7 + x^4 + 3x^2 - x\right) = \lim\limits_{x \rightarrow -\infty} -5x^7 = -5\lim\limits_{x \rightarrow -\infty} x^7
We know that:
\lim\limits_{x \rightarrow -\infty} x^7 = -\infty
Therefore, remembering the negative sign of the coefficient :
\lim\limits_{x \to -\infty} f\left(x\right)=\infty
\lim\limits_{x \to \infty} \left(-x^3 + 4x^2 - 2x + 3\right)
The limit of a polynomial as x approaches infinity or negative infinity depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore we have:
\lim\limits_{x \rightarrow \infty} \left(-x^3 + 4x^2 - 2x + 3\right) = \lim\limits_{x \rightarrow \infty} -x^3 =-\lim\limits_{x \rightarrow \infty} x^3 .
We know that:
\lim\limits_{x \rightarrow \infty} x^3 = \infty
Therefore, since there is a negative sign on the coefficient :
\lim\limits_{x \to \infty} f\left(x\right)=-\infty
\lim\limits_{x \to -\infty} \left(4x^4 + 2x^2 + 6\right)
The limit of a polynomial as x approaches infinity (or negative infinity) depends only on the highest-order term, because at numbers with arbitrarily large magnitudes, the highest-order term dominates.
Therefore we have:
\lim\limits_{x \rightarrow -\infty} \left(4x^4 + 2x^2 + 6\right) = \lim\limits_{x \rightarrow -\infty} 4x^4 = 4\lim\limits_{x \rightarrow -\infty} x^4
We know that:
\lim\limits_{x \rightarrow -\infty} x^4 = \infty
Therefore:
\lim\limits_{x \to -\infty} f\left(x\right)=\infty