Question

2^x-1/√(x+1)-1 evaluate for limit x approaches to zero

Answer 1

Answer:

$\large \bold\red{2 \: ln(2) }$

Step-by-step explanation:

Given that,

• $\lim{x\to 0} \frac{ {2}^{x} - 1}{ \sqrt{(x + 1)} - 1}$

If we put the value of x = 0 in the given limit.

We will get 0/0 form.

Therefore,

We can Apply L- Hospital's Rule here.

• In L - Hospital's Rule, we separately differentiate both Numerator and Denominator till we get a defined value when x = 0 is substituted.

Therefore,

We get,

$= \lim{x\to 0} \frac{ \frac{d}{dx} ( {2}^{x} - 1)}{ \frac{d}{dx}( \sqrt{(x + 1)} - 1) } \\ \\ = \lim{x\to 0} \frac{ \frac{d}{dx} ({2}^{x}) - \frac{d}{dx}(1) }{ \frac{d}{dx} ( \sqrt{x + 1} ) - \frac{d}{dx}(1) }$

But,

We know that,

• $\frac{d}{dx} {a}^{x} = {a}^{x} ln(a)$

• $\frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }$

• $\frac{d}{dx} (constant \: term) = 0$

Therefore,

We get,

$= \lim{x\to 0} \frac{ {2}^{x} ln(2) - 0 }{ \frac{1}{2 \sqrt{x + 1} } - 0 } \\ \\ = \lim{x\to 0} \frac{ {2}^{x} ln(2) }{ \frac{1}{2 \sqrt{x + 1} } } \\ \\ = \frac{ {2}^{0} ln(2) }{ \frac{1}{2 \sqrt{0 + 1} } } \\ \\ = \frac{1 \times ln(2) }{ \frac{1}{2 \sqrt{1} } } \\ \\ = \frac{ ln(2) }{ \frac{1}{2} } \\ \\ = \large \bold{2 \: ln(2) }$