Limits Form 3: Trigonometrical Limits Example 1

Step-by-Step Solution:

As [x→0], we observe:

[\sin 5x → 0]

[3x → 0]
Hence, the limit is of the indeterminate form [\dfrac{0}{0}].

We use the standard trigonometric limit formula:
[\lim_{θ→0} \dfrac{\sin θ}{θ} = 1]

To apply this formula, we rewrite the given expression:
[\dfrac{\sin 5x}{3x} = \dfrac{5}{3} \cdot \dfrac{\sin 5x}{5x}]

As [x→0], [5x→0]. Therefore:
[\dfrac{\sin 5x}{5x} → 1]

Multiplying the constants:
[\dfrac{5}{3} \cdot 1 = \dfrac{5}{3}]

Final Answer:
[\dfrac{5}{3}]

Step-by-Step Solution:

As [x→0]:

[\tan 7x → 0]

[x → 0]
Hence, the form is [\dfrac{0}{0}].

We use the standard result:
[\lim_{θ→0} \dfrac{\tan θ}{θ} = 1]

Rewrite the expression to apply the formula:
[\dfrac{\tan 7x}{x} = 7 \cdot \dfrac{\tan 7x}{7x}]

As [x→0], [7x→0], so:
[\dfrac{\tan 7x}{7x} → 1]

Hence:
[7 \cdot 1 = 7]

Final Answer:
[7]

Step-by-Step Solution:

As [x→0]:

[1-\cos 4x → 0]

[x^{2} → 0]
So the form is [\dfrac{0}{0}].

We use the standard formula:
[\lim_{θ→0} \dfrac{(1-\cos θ)}{(θ^{2})} = \dfrac{1}{2}]

Convert the given expression into standard form:
[\dfrac{1-\cos 4x}{x^{2}} = \dfrac{1-\cos 4x}{(4x)^{2}} \cdot 16]

As [x→0], [4x→0]. Therefore:
[\dfrac{1-\cos 4x}{(4x)^{2}} → \dfrac{1}{2}]

Multiply:
[16 \cdot \dfrac{1}{2} = 8]

Final Answer:
[8]

Step-by-Step Solution:

As [x→0], the expression gives [\dfrac{0}{0}].

Rewrite carefully using standard forms:
[\dfrac{\sin 3x}{\tan 2x} \dfrac{\sin 3x}{3x} \cdot \dfrac{3x}{2x} \cdot \dfrac{2x}{\tan 2x}]

Use standard limits:

[\dfrac{\sin θ}{θ} → 1]

[\dfrac{θ}{\tan θ} → 1]

Hence:
[1 \cdot \dfrac{3}{2} \cdot 1 ][= \dfrac{3}{2}]

Final Answer:
[\dfrac{3}{2}]

Step-by-Step Solution:

As [x→0]:

[\sin 6x → 0]

[4x → 0]
Hence, the form is [\dfrac{0}{0}].

Use the standard limit:
[\lim_{θ→0} \dfrac{\sin θ}{θ} = 1]

Rewrite the expression to match the formula:
[\dfrac{\sin 6x}{4x} \dfrac{6}{4} \cdot \dfrac{\sin 6x}{6x}]

As [x→0], [6x→0], so:
[\dfrac{\sin 6x}{6x} → 1]

Multiply the constants:
[\dfrac{6}{4} = \dfrac{3}{2}]

Final Answer:
[\dfrac{3}{2}]

Step-by-Step Solution:

As [x→0], both numerator and denominator approach zero, so the form is [\dfrac{0}{0}].

Use the standard limit:
[\lim_{θ→0} \dfrac{\tan θ}{θ} = 1]

Rewrite:
[\dfrac{\tan 4x}{5x} = \dfrac{4}{5} \cdot \dfrac{\tan 4x}{4x}]

As [x→0], [4x→0], hence:
[\dfrac{\tan 4x}{4x} → 1]

Multiply:
[\dfrac{4}{5} \cdot 1 = \dfrac{4}{5}]

Final Answer:
[\dfrac{4}{5}]

Step-by-Step Solution:

As [x→0]:

[1-\cos 6x → 0]

[x^{2} → 0]
So the form is [\dfrac{0}{0}].

Use the standard result:
[\lim_{θ→0} \dfrac{(1-\cos θ)}{(θ^{2})} = \dfrac{1}{2}]

Rewrite the expression:
[\dfrac{1-\cos 6x}{x^{2}} = \dfrac{1-\cos 6x}{(6x)^{2}} \cdot 36]

As [x→0], [6x→0], so:
[\dfrac{1-\cos 6x}{(6x)^{2}} → \dfrac{1}{2}]

Multiply:
[36 \cdot \dfrac{1}{2} = 18]

Final Answer:
[18]

Step-by-Step Solution:

As [x→0], the expression is of the form [\dfrac{0}{0}].

Rewrite using standard forms:
[\dfrac{\sin 2x}{\tan 5x} ][= \dfrac{\sin 2x}{2x} \cdot \dfrac{2x}{5x} \cdot \dfrac{5x}{\tan 5x}]

Apply standard limits:

[\dfrac{\sin θ}{θ} → 1]

[\dfrac{θ}{\tan θ} → 1]

Evaluate:
[1 \cdot \dfrac{2}{5} \cdot 1 = \dfrac{2}{5}]

Final Answer:
[\dfrac{2}{5}]

Step-by-Step Solution:

As [x→0], both numerator and denominator tend to zero → form is [\dfrac{0}{0}].

Split the fraction:
[\dfrac{\sin 7x + \tan 3x}{x} ][= \dfrac{\sin 7x}{x} + \dfrac{\tan 3x}{x}]

Rewrite each term:
[\dfrac{\sin 7x}{x} = 7 \cdot \dfrac{\sin 7x}{7x}]
[\dfrac{\tan 3x}{x} = 3 \cdot \dfrac{\tan 3x}{3x}]

Apply standard limits:
[\dfrac{\sin 7x}{7x} → 1]
[\dfrac{\tan 3x}{3x} → 1]

Add the results:
[7 + 3 = 10]

Final Answer:
[10]

Step-by-Step Solution:

As [x→0]:

[\sin^{-1} x → 0]

[x → 0]
Hence, the limit is of the indeterminate form [\dfrac{0}{0}].

We use the standard inverse trigonometric limit:
[\lim_{x→0} \dfrac{\sin^{-1} x}{x} = 1]

Since the given expression is exactly in this form, the value of the limit is directly obtained.

Final Answer:
[1]

Step-by-Step Solution:

As [x→0]:

[\tan^{-1} 5x → 0]

[x → 0]
So, the form is [\dfrac{0}{0}].

Use the standard limit:
[\lim_{θ→0} \dfrac{\tan^{-1} θ}{θ} = 1]

Rewrite the expression:
[\dfrac{\tan^{-1} 5x}{x} ][= 5 \cdot \dfrac{\tan^{-1} 5x}{5x}]

As [x→0], [5x→0], hence:
[\dfrac{\tan^{-1} 5x}{5x} → 1]

Multiply:
[5 \cdot 1 = 5]

Final Answer:
[5]

Step-by-Step Solution:

As [x→0]:

[\sin^{-1} 3x → 0]

[\tan^{-1} 2x → 0]
Hence, the form is [\dfrac{0}{0}].

Rewrite the expression using standard forms:
[\dfrac{\sin^{-1} 3x}{\tan^{-1} 2x} ][= \dfrac{\sin^{-1} 3x}{3x} \cdot \dfrac{3x}{2x} \cdot \dfrac{2x}{\tan^{-1} 2x}]

Apply standard limits:

[\dfrac{\sin^{-1} θ}{θ} → 1]

[\dfrac{θ}{\tan^{-1} θ} → 1]

Evaluate:
[1 \cdot \dfrac{3}{2} \cdot 1 = \dfrac{3}{2}]

Final Answer:
[\dfrac{3}{2}]

Step-by-Step Solution:

As [x→0]:

[\tan^{-1} x → 0]

[\sin x → 0]
So the form is [\dfrac{0}{0}].

Rewrite carefully:
[\dfrac{\tan^{-1} x}{\sin x} ][= \dfrac{\tan^{-1} x}{x} \cdot \dfrac{x}{\sin x}]

Use standard limits:

[\dfrac{\tan^{-1} x}{x} → 1]

[\dfrac{x}{\sin x} → 1]

Multiply:
[1 \cdot 1 = 1]

Final Answer:
[1]

Step-by-Step Solution:

As [x→0], the expression becomes [\dfrac{0}{0}].

Split the fraction:
[\dfrac{\sin^{-1} x + \tan^{-1} x}{x} ][= \dfrac{\sin^{-1} x}{x} + \dfrac{\tan^{-1} x}{x}]

Apply standard limits:

[\dfrac{\sin^{-1} x}{x} → 1]

[\dfrac{\tan^{-1} x}{x} → 1]

Add:
[1 + 1 = 2]

Final Answer:
[2]

Step-by-Step Solution:

As [x→0], the form is [\dfrac{0}{0}].

Rewrite:
[\dfrac{\sin^{-1} 4x}{3x} ][= \dfrac{4}{3} \cdot \dfrac{\sin^{-1} 4x}{4x}]

Apply the standard limit:
[\dfrac{\sin^{-1} θ}{θ} → 1]

Evaluate:
[\dfrac{4}{3} \cdot 1 = \dfrac{4}{3}]

Final Answer:
[\dfrac{4}{3}]