Limits Form 4: Exponential Limits Example 1

Step-by-Step Solution:

As [x→0]:

[\sin x → 0]

[e^{\sin x} → 1]
Hence, the limit is of the form [\dfrac{0}{0}].

Let [u = \sin x].
As [x→0], [u→0].

Rewrite the limit:
[\lim_{u→0} \dfrac{(e^{u}-1)}{u}]

Use the standard exponential limit:
[\lim_{u→0} \dfrac{(e^{u}-1)}{u} = 1]

Final Answer:
[1]

Step-by-Step Solution:

As [x→0]:

[\tan x → 0]

[e^{\tan x} → 1]
So the form is [\dfrac{0}{0}].

Put [u = \tan x].
As [x→0], [u→0].

Convert the limit:
[\lim_{u→0} \dfrac{(e^{u}-1)}{u}]

Apply the standard result.

Final Answer:
[1]

Step-by-Step Solution:

As [x→0], the form is [\dfrac{0}{0}].

Rewrite:
[\dfrac{e^{2\sin x}-1}{\sin x} ][= 2 \cdot \dfrac{e^{2\sin x}-1}{2\sin x}]

Let [u = 2\sin x].
As [x→0], [u→0].

Apply the standard limit:
[\dfrac{e^{u}-1}{u} → 1]

Multiply:
[2 \cdot 1 = 2]

Final Answer:
[2]

Step-by-Step Solution:

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

Rewrite:
[\dfrac{e^{3\tan x}-1}{\tan x} ][= 3 \cdot \dfrac{e^{3\tan x}-1}{3\tan x}]

Let [u = 3\tan x], then [u→0].

Apply:
[\dfrac{e^{u}-1}{u} → 1]

Multiply:
[3]

Final Answer:
[3]

Step-by-Step Solution:

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

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

Apply limits separately:

[\dfrac{e^{\sin x}-1}{\sin x} → 1]

[\dfrac{\sin x}{x} → 1]

Multiply:
[1 \cdot 1 = 1]

Final Answer:
[1]

Step-by-Step Solution:

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

Apply limits:

[\dfrac{e^{\tan x}-1}{\tan x} → 1]

[\dfrac{\tan x}{x} → 1]

Multiply:
[1]

Final Answer:
[1]

Step-by-Step Solution:

Rewrite carefully:
[\dfrac{e^{2\sin x}-1}{x} ][= 2 \cdot \dfrac{e^{2\sin x}-1}{2\sin x} \cdot \dfrac{\sin x}{x}]

Apply limits:

[\dfrac{e^{2\sin x}-1}{2\sin x} → 1]

[\dfrac{\sin x}{x} → 1]

Multiply:
[2]

Final Answer:
[2]

Step-by-Step Solution:

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

Apply limits:

[\dfrac{e^{5\tan x}-1}{5\tan x} → 1]

[\dfrac{\tan x}{x} → 1]

Multiply:
[5]

Final Answer:
[5]

Step-by-Step Solution:

Rewrite:
[\dfrac{e^{\sin 2x}-1}{x} ][= \dfrac{e^{\sin 2x}-1}{\sin 2x} \cdot \dfrac{\sin 2x}{2x} \cdot 2]

Apply limits:

[\dfrac{e^{\sin 2x}-1}{\sin 2x} → 1]

[\dfrac{\sin 2x}{2x} → 1]

Multiply:
[2]

Final Answer:
[2]

Step-by-Step Solution:

Rewrite:
[\dfrac{e^{\tan 3x}-1}{x} ][= \dfrac{e^{\tan 3x}-1}{\tan 3x} \cdot \dfrac{\tan 3x}{3x} \cdot 3]

Apply limits:

[\dfrac{e^{\tan 3x}-1}{\tan 3x} → 1]

[\dfrac{\tan 3x}{3x} → 1]

Multiply:
[3]

Final Answer:
[3]