Limit Evaluation: X->0 (e^(1+xsinx)-e^(cos2x))/(√(1+x^2)-1)

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Hey guys! Let's dive into a fascinating limit problem today. We're going to tackle the evaluation of the following limit as x approaches 0:

lim (x->0) [ (e^(1 + x*sin(x)) - e^(cos(2x))) / (sqrt(1 + x^2) - 1) ]

This looks pretty intense, but don't worry, we'll break it down step by step. Grab your thinking caps, and let's get started!

Unpacking the Problem

Before we even start crunching numbers, it's super important to understand what we're dealing with. Limit problems like this one are all about figuring out what happens to a function as its input (in this case, x) gets closer and closer to a specific value (here, 0).

Our function is a big fraction, and both the numerator and the denominator look a bit scary. The numerator involves exponential functions (e raised to some power) and trigonometric functions (sin and cos). The denominator has a square root, which often hints at needing some algebraic manipulation. This means that our key strategy involves understanding exponential behavior, trigonometric identities, and some clever algebraic tricks to simplify the expression. The main keyword here is understanding the limit behavior near zero which will make solving this question easier.

Initial Assessment

Our very first step should always be to try plugging in x = 0 directly into the expression. This gives us:

(e^(1 + 0*sin(0)) - e^(cos(2*0))) / (sqrt(1 + 0^2) - 1)
= (e^(1 + 0) - e^(cos(0))) / (sqrt(1) - 1)
= (e^1 - e^1) / (1 - 1)
= 0 / 0

Uh oh! We've landed on the indeterminate form 0/0. This doesn't mean the limit doesn't exist, but it does mean we need to do some more work. It’s a clear signal that we need to manipulate the expression somehow to get it into a form where we can evaluate the limit. L'Hôpital's Rule might come to mind, but let's first explore if we can solve this using series expansions and algebraic manipulations, which often give more insight.

Leveraging Series Expansions

When dealing with limits, especially around x = 0, series expansions are your best friends. They allow us to approximate functions using polynomials, which are much easier to work with. The most useful expansions for our problem are the Maclaurin series, which are Taylor series centered at 0.

Key Series Expansions

Let's recall some crucial Maclaurin series that we'll need:

  1. Exponential Function:

    e^u = 1 + u + u^2/2! + u^3/3! + ...

  2. Sine Function:

    sin(x) = x - x^3/3! + x^5/5! - ...

  3. Cosine Function:

    cos(x) = 1 - x^2/2! + x^4/4! - ...

  4. Square Root Function: Using the binomial series expansion for (1 + u)^(1/2):

    sqrt(1 + u) = 1 + (1/2)u - (1/8)u^2 + (1/16)u^3 - ...

These expansions are the fundamental tools we’ll use to simplify our limit expression. Recognizing when and how to apply these is a critical skill in calculus.

Applying the Expansions

Now, let's apply these series expansions to the functions in our limit.

Expanding the Numerator

We have e^(1 + x*sin(x)) - e^(cos(2x)). Let's tackle each exponential term separately.

First, consider e^(1 + x*sin(x)). We can rewrite this as e * e^(x*sin(x)). Now, let's focus on e^(x*sin(x)). Using the series expansion for e^u, where u = x*sin(x):

e^(x*sin(x)) = 1 + x*sin(x) + (x*sin(x))^2/2! + ...

Next, we need to expand sin(x) itself:

sin(x) = x - x^3/3! + O(x^5)

So,

x*sin(x) = x*(x - x^3/3! + ...) = x^2 - x^4/6 + O(x^6)

Plugging this back into the exponential series:

e^(x*sin(x)) = 1 + (x^2 - x^4/6) + (x^2 - x^4/6)^2/2 + O(x^6)
             = 1 + x^2 - x^4/6 + x^4/2 + O(x^6)
             = 1 + x^2 + x^4/3 + O(x^6)

Thus,

e^(1 + x*sin(x)) = e * (1 + x^2 + x^4/3 + O(x^6))
                    = e + ex^2 + ex^4/3 + O(x^6)

Now, let's expand e^(cos(2x)). Using the series expansion for e^u, where u = cos(2x):

e^(cos(2x)) = 1 + cos(2x) + cos^2(2x)/2! + ...

We need the series expansion for cos(2x):

cos(2x) = 1 - (2x)^2/2! + (2x)^4/4! - ...
        = 1 - 2x^2 + 2x^4/3 - O(x^6)

So, let's substitute cos(2x) in the expansion for e^u:

e^(cos(2x)) = e^(1 - 2x^2 + 2x^4/3 + O(x^6))
            = e * e^(-2x^2 + 2x^4/3 + O(x^6))

Now, let v = -2x^2 + 2x^4/3 + O(x^6) and expand e^v:

e^v = 1 + v + v^2/2! + ...
    = 1 + (-2x^2 + 2x^4/3) + (-2x^2 + 2x^4/3)^2/2 + O(x^6)
    = 1 - 2x^2 + 2x^4/3 + 2x^4 + O(x^6)
    = 1 - 2x^2 + 8x^4/3 + O(x^6)

Therefore,

e^(cos(2x)) = e * (1 - 2x^2 + 8x^4/3 + O(x^6))
            = e - 2ex^2 + 8ex^4/3 + O(x^6)

Now we subtract the two expanded terms:

e^(1 + x*sin(x)) - e^(cos(2x)) = (e + ex^2 + ex^4/3) - (e - 2ex^2 + 8ex^4/3) + O(x^6)
                             = 3ex^2 - 7ex^4/3 + O(x^6)

Expanding the Denominator

Our denominator is sqrt(1 + x^2) - 1. We can use the binomial series expansion for sqrt(1 + u) with u = x^2:

sqrt(1 + x^2) = 1 + (1/2)x^2 - (1/8)x^4 + O(x^6)

So,

sqrt(1 + x^2) - 1 = (1/2)x^2 - (1/8)x^4 + O(x^6)

Reassembling the Limit

Now, let's put the expanded numerator and denominator back into our limit:

lim (x->0) [ (e^(1 + x*sin(x)) - e^(cos(2x))) / (sqrt(1 + x^2) - 1) ]
= lim (x->0) [ (3ex^2 - 7ex^4/3 + O(x^6)) / ((1/2)x^2 - (1/8)x^4 + O(x^6)) ]

We can divide both the numerator and the denominator by x^2:

lim (x->0) [ (3e - 7ex^2/3 + O(x^4)) / (1/2 - (1/8)x^2 + O(x^4)) ]

As x approaches 0, the higher-order terms (x^2, x^4, etc.) go to 0, so we are left with:

lim (x->0) [ 3e / (1/2) ] = 6e

Final Answer

Therefore, the limit is:

lim (x->0) [ (e^(1 + x*sin(x)) - e^(cos(2x))) / (sqrt(1 + x^2) - 1) ] = 6e

So, guys, we've successfully navigated this tricky limit problem by using series expansions and algebraic manipulation. The key takeaway here is that series expansions are a powerful tool when dealing with limits of complicated functions, especially around x = 0. Remember to identify the relevant expansions, apply them carefully, and simplify the expression to reveal the limit. Hope this was helpful, and happy problem-solving!