# FIND THE EQUATION OF THE TANGENT LINE USING LIMITS

## About "Find the equation of the tangent line using limits"

Find the equation of the tangent line using limits :

The line which passes though any point lies on the curve is known as tangent line.To find the equation of any line, we need two information.

(i) Point on the line (The point is common both line and curve)

(ii) Slope of tangent

Example 1 :

Find the equations of the tangent to the curve y = x² – x – 2 at x = 1

Solution :

To find the point, first let us apply the value of x in the given function

y  =  1² – 1 – 2

=  1  - 1 - 2

=  -2

The required point lies on the tangent line and the curve is (1, -2).

Now we need to find the slope of the tangent. For that we have to find the derivative of the given curve.

y = x² – x – 2

dy/dx  =  2x - 1

Slope at x = 1

dy/dx  =  2(1) - 1  ==>  2 - 1 =  1 ==> m  =  1

Equation of the tangent :

(y - y₁) = m (x - x₁)

Here (x₁, y₁) ==> (1, -2) and m = 1

(y - (-2))  =  1 (x - 1)

(y + 2) = 1 (x - 1)

y + 2 = x - 1

Subtract y and 2 on both sides

y + 2 - y - 2 = x - 1 - y - 2

0 = x - y - 3

x - y - 3 = 0

Hence the required equation of the tangent line is x - y - 3 = 0.

Example 2 :

Find the equations of the tangent to the curve y = x² – 4x – 5 at x = -2

Solution :

y = x² – 4x – 5

=  (-2)² – 4(-2) – 5

=  4 + 8 - 5 ==> 12 - 5 ==> 7

The required point lies on the tangent line and the curve is (-2, 7).

Now we need to find the slope of the tangent. For that we have to find the derivative of the given curve.

y = x² – 4x – 5

dy/dx  =  2x - 4

Slope at x = -2

dy/dx  =  2(-2) - 4

=  -4 - 4

= -8

Hence the required slope (m)  =  -8

Equation of the tangent :

(y - y₁) = m (x - x₁)

Here (x₁, y₁) ==> (-2, 7) and m = -8

(y - 7)  =  1 (x - (-2))

(y - 7) = 1 (x + 2)

y - 7 = x + 2

Subtract y and add 7 on both sides

y - 7 - y + 7  = x + 2 - y + 7

0 = x - y + 9

x - y + 9 = 0

Hence the required equation of the tangent line is x - y + 9 = 0.

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