Limits Form 3: Trigonometrical Limits Example 2

Step-by-Step Solution:

As [x→0]:

[\cos 3x → 1]

Hence, [1-\cos 3x → 0]

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

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

Convert the given expression into standard form:
[\dfrac{1-\cos 3x}{x^{2}} ][= \dfrac{1-\cos 3x}{(3x)^{2}} \cdot 9]

As [x→0], [3x→0], hence:
[\dfrac{1-\cos 3x}{(3x)^{2}} → \dfrac{1}{2}]

Multiply:
[9 \cdot \dfrac{1}{2} = \dfrac{9}{2}]

Final Answer:
[\dfrac{9}{2}]

Step-by-Step Solution:

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

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

Rewrite:
[\dfrac{1-\cos 5x}{x^{2}} ][= \dfrac{1-\cos 5x}{(5x)^{2}} \cdot 25]

Taking the limit:
[\dfrac{1-\cos 5x}{(5x)^{2}} → \dfrac{1}{2}]

Final calculation:
[25 \cdot \dfrac{1}{2} = \dfrac{25}{2}]

Final Answer:
[\dfrac{25}{2}]

Step-by-Step Solution:

As [x→0], numerator and denominator both tend to zero.

Factor the denominator:
[4x^{2} = (2x)^{2}]

Rewrite:
[\dfrac{1-\cos 2x}{(2x)^{2}}]

Apply the standard result:
[\dfrac{1-\cos θ}{θ^{2}} → \dfrac{1}{2}]

Hence:
[\dfrac{1}{2}]

Final Answer:
[\dfrac{1}{2}]

Step-by-Step Solution:

As [x→0]:

[\cos 7x → 1]

Therefore, [1-\cos 7x → 0]

[49x^{2} → 0]
Hence, the limit is of the indeterminate form [\dfrac{0}{0}].

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

Rewrite the denominator:
[49x^{2} = (7x)^{2}]

Rewrite the given expression:
[\dfrac{1 – \cos 7x}{49x^{2}} = \dfrac{1 – \cos 7x}{(7x)^{2}}]

As [x→0], [7x→0], so:
[\dfrac{1-\cos 7x}{(7x)^{2}} → \dfrac{1}{2}]

Final Answer:
[\dfrac{1}{2}]

Step-by-Step Solution:

As [x→0]:

[1-\cos 4x → 0]

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

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

Rewrite the expression:
[\dfrac{1 – \cos 4x}{2x^{2}} ][= \dfrac{1}{2} \cdot \dfrac{1 – \cos 4x}{x^{2}}]

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

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

Multiply step by step:
[\dfrac{1}{2} \cdot 16 \cdot \dfrac{1}{2} = 4]

Final Answer:
[4]

Step-by-Step Solution:

As [x→0]:

[1-\cos x → 0]

[3x^{2} → 0]
So the form is [\dfrac{0}{0}].

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

Rewrite:
[\dfrac{1-\cos x}{3x^{2}} ][= \dfrac{1}{3} \cdot \dfrac{1-\cos x}{x^{2}}]

Taking the limit:
[\dfrac{1-\cos x}{x^{2}} → \dfrac{1}{2}]

Multiply:
[\dfrac{1}{3} \cdot \dfrac{1}{2} = \dfrac{1}{6}]

Final Answer:
[\dfrac{1}{6}]

Step-by-Step Solution:

As [x→0], the form is [\dfrac{0}{0}].

Rewrite the expression:
[\dfrac{1-\cos 6x}{9x^{2}} ][= \dfrac{1-\cos 6x}{(6x)^{2}} \cdot \dfrac{36}{9}]

Apply standard limit:
[\dfrac{1-\cos 6x}{(6x)^{2}} → \dfrac{1}{2}]

Simplify:
[\dfrac{36}{9} = 4]

Multiply:
[\dfrac{1}{2} \cdot 4 = 2]

Final Answer:
[2]

Step-by-Step Solution:

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

Rewrite the denominator:
[16x^{2} = (4x)^{2}]

Rewrite the expression:
[\dfrac{1-\cos 8x}{(4x)^{2}} ][= \dfrac{1-\cos 8x}{(8x)^{2}} \cdot 4]

Apply standard limit:
[\dfrac{1-\cos 8x}{(8x)^{2}} → \dfrac{1}{2}]

Multiply:
[\dfrac{1}{2} \cdot 4 = 2]

Final Answer:
[2]

Step-by-Step Solution:

As [x→0], form is [\dfrac{0}{0}].

Rewrite:
[\dfrac{1-\cos 10x}{5x^{2}} ][= \dfrac{1-\cos 10x}{(10x)^{2}} \cdot \dfrac{100}{5}]

Apply standard limit:
[\dfrac{1-\cos 10x}{(10x)^{2}} → \dfrac{1}{2}]

Simplify:
[\dfrac{100}{5} = 20]

Multiply:
[\dfrac{1}{2} \cdot 20 = 10]

Final Answer:
[10]

Step-by-Step Solution:

As [x→0], the limit is [\dfrac{0}{0}].

Rewrite:
[\dfrac{1-\cos 12x}{6x^{2}} ][= \dfrac{1-\cos 12x}{(12x)^{2}} \cdot \dfrac{144}{6}]

Apply standard limit:
[\dfrac{1-\cos 12x}{(12x)^{2}} → \dfrac{1}{2}]

Simplify:
[\dfrac{144}{6} = 24]

Multiply:
[\dfrac{1}{2} \cdot 24 = 12]

Final Answer:
[12]