Calculus Examples

$lim_{x \to 1} \frac{2 x - 7}{4 x + 5}$

Step 1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1} 2 x - 7}{lim_{x \to 1} 4 x + 5}$

Step 2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1} 2 x - lim_{x \to 1} 7}{lim_{x \to 1} 4 x + 5}$

Step 3

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

$\frac{2 lim_{x \to 1} x - lim_{x \to 1} 7}{lim_{x \to 1} 4 x + 5}$

Step 4

Evaluate the limit of $7$ which is constant as $x$ approaches $1$ .

$\frac{2 lim_{x \to 1} x - 1 \cdot 7}{lim_{x \to 1} 4 x + 5}$

Step 5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{2 lim_{x \to 1} x - 1 \cdot 7}{lim_{x \to 1} 4 x + lim_{x \to 1} 5}$

Step 6

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 lim_{x \to 1} x - 1 \cdot 7}{4 lim_{x \to 1} x + lim_{x \to 1} 5}$

Step 7

Evaluate the limit of $5$ which is constant as $x$ approaches $1$ .

$\frac{2 lim_{x \to 1} x - 1 \cdot 7}{4 lim_{x \to 1} x + 5}$

Step 8

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 8.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{2 \cdot 1 - 1 \cdot 7}{4 lim_{x \to 1} x + 5}$

Step 8.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{2 \cdot 1 - 1 \cdot 7}{4 \cdot 1 + 5}$

Step 9

Simplify the answer.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Multiply $2$ by $1$ .

$\frac{2 - 1 \cdot 7}{4 \cdot 1 + 5}$

Step 9.1.2

Multiply $- 1$ by $7$ .

$\frac{2 - 7}{4 \cdot 1 + 5}$

Step 9.1.3

Subtract $7$ from $2$ .

$\frac{- 5}{4 \cdot 1 + 5}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Multiply $4$ by $1$ .

$\frac{- 5}{4 + 5}$

Step 9.2.2

Add $4$ and $5$ .

$\frac{- 5}{9}$

Step 9.3

Move the negative in front of the fraction.

$- \frac{5}{9}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$- \frac{5}{9}$

Decimal Form:

$- 0.\overline{5}$