Calculus Examples

$lim_{x \to - \infty} \frac{x^{2}}{e^{1 - x}}$

Step 1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to - \infty} x^{2}}{lim_{x \to - \infty} e^{1 - x}}$

Step 1.2

The limit at negative infinity of a polynomial of even degree whose leading coefficient is positive is infinity.

$\frac{\infty}{lim_{x \to - \infty} e^{1 - x}}$

Step 1.3

Since the exponent $1 - x$ approaches $\infty$ , the quantity $e^{1 - x}$ approaches $\infty$ .

$\frac{\infty}{\infty}$

Step 1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 2

Since $\frac{\infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to - \infty} \frac{x^{2}}{e^{1 - x}} = lim_{x \to - \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{1 - x}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{1 - x}]}$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to - \infty} \frac{2 x}{\frac{d}{d x} [e^{1 - x}]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 1 - x$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $1 - x$ .

$lim_{x \to - \infty} \frac{2 x}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [1 - x]}$

Step 3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$lim_{x \to - \infty} \frac{2 x}{e^{u} \frac{d}{d x} [1 - x]}$

Step 3.3.3

Replace all occurrences of $u$ with $1 - x$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} \frac{d}{d x} [1 - x]}$

Step 3.4

By the Sum Rule, the derivative of $1 - x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x]$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} (\frac{d}{d x} [1] + \frac{d}{d x} [- x])}$

Step 3.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} (0 + \frac{d}{d x} [- x])}$

Step 3.6

Add $0$ and $\frac{d}{d x} [- x]$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} \frac{d}{d x} [- x]}$

Step 3.7

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} (- \frac{d}{d x} [x])}$

Step 3.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} (- 1 \cdot 1)}$

Step 3.9

Multiply $- 1$ by $1$ .

$lim_{x \to - \infty} \frac{2 x}{e^{1 - x} \cdot - 1}$

Step 3.10

Move $- 1$ to the left of $e^{1 - x}$ .

$lim_{x \to - \infty} \frac{2 x}{- 1 e^{1 - x}}$

Step 3.11

Rewrite $- 1 e^{1 - x}$ as $- e^{1 - x}$ .

$lim_{x \to - \infty} \frac{2 x}{- e^{1 - x}}$

Step 4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to - \infty} \frac{x}{- e^{1 - x}}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 5.1.1

Take the limit of the numerator and the limit of the denominator.

$2 \frac{lim_{x \to - \infty} x}{lim_{x \to - \infty} - e^{1 - x}}$

Step 5.1.2

The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.

$2 \frac{- \infty}{lim_{x \to - \infty} - e^{1 - x}}$

Step 5.1.3

Since the function $e^{1 - x}$ approaches $\infty$ , the negative constant $- 1$ times the function approaches $- \infty$ .

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Step 5.1.3.1

Consider the limit with the constant multiple $- 1$ removed.

$2 \frac{- \infty}{lim_{x \to - \infty} e^{1 - x}}$

Step 5.1.3.2

Since the exponent $1 - x$ approaches $\infty$ , the quantity $e^{1 - x}$ approaches $\infty$ .

$2 \frac{- \infty}{\infty}$

Step 5.1.3.3

Since the function $e^{1 - x}$ approaches $\infty$ , the negative constant $- 1$ times the function approaches $- \infty$ .

$2 \frac{- \infty}{- \infty}$

Step 5.1.3.4

Infinity divided by infinity is undefined.

Undefined

$2 \frac{- \infty}{- \infty}$

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

$2 \frac{- \infty}{- \infty}$

Step 5.2

Since $\frac{- \infty}{- \infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to - \infty} \frac{x}{- e^{1 - x}} = lim_{x \to - \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [- e^{1 - x}]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$2 lim_{x \to - \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [- e^{1 - x}]}$

Step 5.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 lim_{x \to - \infty} \frac{1}{\frac{d}{d x} [- e^{1 - x}]}$

Step 5.3.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{1 - x}$ with respect to $x$ is $- \frac{d}{d x} [e^{1 - x}]$ .

$2 lim_{x \to - \infty} \frac{1}{- \frac{d}{d x} [e^{1 - x}]}$

Step 5.3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 1 - x$ .

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Step 5.3.4.1

To apply the Chain Rule, set $u$ as $1 - x$ .

$2 lim_{x \to - \infty} \frac{1}{- \frac{d}{d u} [e^{u}] \frac{d}{d x} [1 - x]}$

Step 5.3.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{u} \frac{d}{d x} [1 - x]}$

Step 5.3.4.3

Replace all occurrences of $u$ with $1 - x$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{1 - x} \frac{d}{d x} [1 - x]}$

Step 5.3.5

By the Sum Rule, the derivative of $1 - x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x]$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{1 - x} (\frac{d}{d x} [1] + \frac{d}{d x} [- x])}$

Step 5.3.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{1 - x} (0 + \frac{d}{d x} [- x])}$

Step 5.3.7

Add $0$ and $\frac{d}{d x} [- x]$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{1 - x} \frac{d}{d x} [- x]}$

Step 5.3.8

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$2 lim_{x \to - \infty} \frac{1}{- e^{1 - x} \cdot - 1 \frac{d}{d x} [x]}$

Step 5.3.9

Multiply $- 1$ by $- 1$ .

$2 lim_{x \to - \infty} \frac{1}{1 e^{1 - x} \frac{d}{d x} [x]}$

Step 5.3.10

Multiply $e^{1 - x}$ by $1$ .

$2 lim_{x \to - \infty} \frac{1}{e^{1 - x} \frac{d}{d x} [x]}$

Step 5.3.11

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 lim_{x \to - \infty} \frac{1}{e^{1 - x} \cdot 1}$

Step 5.3.12

Multiply $e^{1 - x}$ by $1$ .

$2 lim_{x \to - \infty} \frac{1}{e^{1 - x}}$

Step 6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{1 - x}}$ approaches $0$ .

$2 \cdot 0$

Step 7

Multiply $2$ by $0$ .

$0$