By using the following standard limits we will have extra advantage.
\[ \lim_{x\to0} \frac{\sin x}{x} = 1 \]
\[ \lim_{x\to0} \frac{\tan x}{x} = 1 \]
\[ \lim_{x\to0} \frac{\ln(1+x)}{x} = 1 \]
\[ \lim_{x\to0} \frac{e^{x}-1}{x} = 1 \]
While applying these standard limits, make sure that the “standard limit part” (for example \( \frac{\sin x}{x} \), \( \frac{\tan x}{x} \), \( \frac{\ln(1+x)}{x} \), \( \frac{e^{x}-1}{x} \)) is joined to the remaining part of the function by a product symbol — like a factor in \( \big(\frac{\sin x}{x}\big)\cdot g(x) \).
Do NOT treat these as standard limits when that part is:
- inside a logarithm or another function (e.g. \( \ln\!\big(\frac{\sin x}{x}\big) \))
- inside a bigger expression or power (e.g. \( \big(\frac{\sin x}{x} + 1\big)^{2} \))
- added or subtracted with other terms (e.g. \( \frac{\sin x}{x} + 3 \))
These standard limits are valid only when the limit expression appears as a separate factor (connected by a product sign) with the rest of the function.
PROBLEM 4
Evaluate \( \displaystyle \lim_{x\to0} \frac{2x}{(1 - x^{2})\tan x} \)
Given
\[ \lim_{x\to0} \frac{2x}{(1 - x^{2})\tan x} \quad \left( \frac{0}{0} \right) \]
\[ = \lim_{x\to0} \frac{2x}{(1 - x^{2})\tan x} \]
\[ = \lim_{x\to0} \frac{2}{1 - x^{2}} \cdot \frac{x}{\tan x} \]
\[ = \lim_{x\to0} \frac{2}{1 - x^{2}} \cdot 1 \]
(since the standard limit \( \displaystyle \lim_{x\to0} \frac{\tan x}{x} = 1 \) appears as a product w.r.t the remaining part)
\[ = \frac{2}{1 - x^{2}} \]
\[ = \frac{2}{1 - 0^{2}} \]
\[ = 2 \]
PROBLEM 5
Evaluate \( \displaystyle \lim_{x\to0} \frac{\log(1 - x^{2})}{\log(\cos x)} \)
Given
\[ \lim_{x\to0} \frac{\log(1 - x^{2})}{\log(\cos x)} \quad \left( \frac{0}{0} \right) \]
\[ = \lim_{x\to0} \frac{\frac{-2x}{1 - x^{2}}}{\frac{-\sin x}{\cos x}} \]
(Applying L'Hospital's Rule)
\[ = \lim_{x\to0} \frac{-2x}{1 - x^{2}} \times \frac{\cos x}{-\sin x} \]
\[ = \lim_{x\to0} \frac{2x}{1 - x^{2}} \times \frac{\cos x}{\sin x} \]
\[ = \lim_{x\to0} 2 \times \frac{x}{\sin x} \times \frac{\cos x}{1 - x^{2}} \]
(since \( \displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1 \))
\[ = 2 \times 1 \times 1 \]
\[ = 2 \]
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PROBLEM 6
Evaluate \( \displaystyle \lim_{x\to0} \frac{\log(1 - x^{2})}{\log(\cos x)} \)
Given
\[ \lim_{x\to0} \frac{\log(1 - x^{2})}{\log(\cos x)} \quad \left( \frac{0}{0} \right) \]
\[ = \lim_{x\to0} \frac{\frac{-2x}{1 - x^{2}}}{\frac{-\sin x}{\cos x}} \]
(Applying L'Hospital's Rule)
\[ = \lim_{x\to0} \frac{-2x}{1 - x^{2}} \times \frac{\cos x}{-\sin x} \]
\[ = \lim_{x\to0} \frac{2x}{1 - x^{2}} \times \frac{\cos x}{\sin x} \]
\[ = \lim_{x\to0} 2 \times \frac{x}{\sin x} \times \frac{\cos x}{1 - x^{2}} \]
(since \( \displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1 \))
\[ = 2 \times 1 \times 1 \]
\[ = 2 \]
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