Problem 236 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x^{2}}\\{\text { Ans. }+\infty,-\infty\\} $$

Step-by-Step Solution

Answer

The limit of the given function as x approaches 0 is: \[\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x^{2}} = +\infty\]

Step 1: Identify the function and the value x is approaching

We are given the function: \[\frac{\sqrt{1+x}-1}{x^{2}}\] and we need to find the limit as x approaches 0.

Step 2: Direct substitution

Directly substitute x=0 in the function to see if there's a direct output: \[\frac{\sqrt{1+0}-1}{0^{2}} = \frac{\sqrt{1}-1}{0}\] Since we have a division by 0, we cannot directly find the limit this way.

Step 3: Try rationalization

Rationalization is a technique used to simplify expressions involving square roots. Multiply the numerator and denominator with the conjugate of the numerator:\[\frac{\sqrt{1+x}-1}{x^{2}} \cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\]

Step 4: Perform multiplication

Now, we will multiply the numerators and the denominators separately:\[\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x^{2}(\sqrt{1+x}+1)}\]

Step 5: Simplify the numerator

Simplify the numerator using the difference of squares formula:\[\frac{(1+x)-1}{x^{2}(\sqrt{1+x}+1)}\]

Step 6: Simplify the expression further

After simplifying the numerator, we are left with the expression: \[\frac{x}{x^{2}(\sqrt{1+x}+1)}\] Now, cancel out one x term from both the numerator and the denominator:\[\frac{1}{x(\sqrt{1+x}+1)}\]

Step 7: Direct substitution again

Now that we have simplified the expression, let's try direct substitution again, by substituting x=0: \[\frac{1}{0(\sqrt{1+0}+1)} = \frac{1}{0(2)}\] We still have a division by 0.

Step 8: Analyze the function as x approaches 0

Since direct substitution is still giving a division by 0, we need to analyze the behavior of the function as x approaches 0. We know that x is squared in the denominator, so the denominator is always positive. The numerator is always positive, so the function is always positive. Now, as x approaches 0 from the positive side, the denominator gets smaller and smaller, which means the value of the function becomes larger and larger. Similarly, as x approaches 0 from the negative side, the value of the function also becomes larger. Thus, we can conclude that the limit of the given function as x approaches 0 is: \[\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x^{2}} = +\infty\]